NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics – PDF Download

Ans Given-

Initial concentration (R₁) = 0.03M

Final concentration (R₂) = 0.02M

Time = 25 mins.

The formula for average rate of the reaction is,

r_av = - Δ{R} / Δt → Equation 1

∵ {R} = (R₂)-( R₁), the equation 1 is written as,

r_av = - R₂ - R₁ / Δt

= -0.02-0.03 / 25

= 4 X 10^-4 mol L^-1 min^-1

The average rate of reaction in seconds is given by,

= 4 X 10^-4 mol L^-1 / 60 S

(dividing by 60 to convert minutes to seconds)

= 6.6 × 10^-6 mol L^-1 s^-1

The average rate of the reaction in minutes is 4 × 10^-4 mol L^-1 min^-1 and in seconds is 6.6 × 10^-6 mol L^-1 s^-1

Ans Given-

Initial concentration (A₁) = 0.5M

Final concentration (A₂) = 0.4M

Time = 10 mins.

The formula for average rate of the reaction is,

r_av = -1/2 X Δ{A} / Δt → Equation 1

∵ {A} = (A₂)-( A₁), the equation 1 is written as,

r_av = -1/2 X A₂ - A₁ / Δt

= -1/2 X 0.4-0.5 / 10

= -1/2 X 0.1 / 10

= 0.005 mol L^-1 min^-1

= 5 × 10^-3 mol L^-1 min^-1

The average rate of the reaction is 5 × 10^-3 mol L^-1 min^-1

Ans The order of the reaction is sum of the powers of concentration of reactants in the rate law. According to this,

The order of the reaction = 1/2 + 2

= 2 1/2 or 2.5

The order of the reaction is 2.5

Ans Let the reaction be X→Y

As this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)² → Equation 1

Since the concentration of X is increased three times, Equation 1 will become,

Rate = k(3X)²

= k × 9(X)²

∴ The rate of formation of Y will become 9 times.

Thus, the rate of formation of Y will become 9 times

Ans Given-

Rate constant, k = 1.15 × 10^-3 s^-1

innitial quantity R₀ = 5g

Final quantity, R = 3 g

According to the formula of first order reaction

K = 2.303/ t log R₀ / R

1.15 X 10^-3  = 2.303 /t log 5/3

t = 2.303 / 1.15 X 10^-3 log 5/3

t = 443.8 s

t = 4.438 × 10² s

The time taken for 5g of the reactant to reduce to 3 g is 4.438 × 10² s

Ans Given-

t1/2 = 60 mins

Using the formula for half life, t1/2  = 0.693/k , we get,

k= 0.693 / t1/2

k= 0.693 / 60

∴k = 1.155 × 10^-2 min^-1

The rate constant of the reaction, k is 1.155 × 10^-2 min^-1

Ans The rate constant of a chemical reaction normally increases with increase in temperature. It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled and this temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation,

k = A

Thus, the rate constant of a chemical reaction normally increases with increase in temperature.

Ans Given-

Initial temperature, T₁ = 298 K

Final temperature, T₂ = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k₁ = k

Final value of rate constant, k₂ = 2k

Using Arrhenius equation,

 Ans Given-

Energy of activation, E_a= 209.5 kJ mol^–1

E_a in joules = 209.5 × 1000 = 209500 J mol^–1

Temperature, T = 581K

Gas constant, R = 8.314 J K^-1 mol^-1

Using the Arrhenius equation,

k = Ae^-Ea/RT

In this equation, the term e^-Ea/RT represents the number of molecules which have kinetic energy greater than the activation energy, E_a.

∴The number of molecules = e^-Ea/RT → Equation 1. Substituting all the known values in equation 1, we get,

= e^{-209500/8.314X581}

=e^-43.3708

Finding the value of anti ln(43.3708), we get, 1.47 × 10^-19

The fraction of molecules of reactants having energy equal to or greater than activation energy is 1.47 × 10^-19

Exercise

Q 4.1 From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3NO(g) → N2O (g) Rate = k[NO]2

(ii) H2O2 (aq) + 3I– (aq) + 2H+ → 2H2O (l) + 3 I − Rate = k[H2O2 ][I- ]

(iii) CH3CHO (g) → CH4 (g) + CO(g) Rate = k [CH3CHO]3/2

(iv) C2H5Cl (g) → C2H4 (g) + HCl (g) Rate = k [C2H5Cl]

Ans

(i) 3NO(g) → N₂O(g)

Rate = k[NO]²

Here rate law is dependent on concentration of NO only. And power raised to concentration of NO is 2. Hence order of reaction is 2.

To find the dimensions of rate constant (K)- We know the rate of chemical reaction is measured in (concentration / time). So, the unit of K will be

(concentration) / (time) (concentration)²

 i.e. [concentration]^-1[ time]^-1 . If concentration is in mol L^-1 and time is in sec. then dimensions of k will be : mol^-1 L sec^-1

(ii) H₂O₂(aq) + 3I^-(aq) + 2H⁺ → 2 H₂O(l) + I₃^-

Rate = k[H₂O₂][I^-]

Here rate law is dependent on concentration of H₂O₂ and I^-. And power raised to concentration of H₂O₂ is 1, and power raised to concentration of I^- is 1. order is sum of powers hence, order of given reaction is 2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]² i.e. [concentration]^-1[ time]^-1. If concentration is in mol L^-1 and time is in sec then dimensions of k will be mol^-1 L sec^-1

(iii) CH₃CHO(g) → CH₄(g) + CO(g)

Rate = k[CH₃CHO]^3/2

Here rate law is dependent on concentration of CH₃CHO.And power raised to CH₃CHO is 3/2. Hence order of reaction is 3/2.

To find dimensions of rate constant (K)- We know rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]^3/2 i.e. [concentration]^-

^1/2[time]^-1.

If concentration is in mol L^-1 and time is in sec. Then dimensions of K will be mol^-1/2L^1/2 sec^-1

(iv) C₂H₅Cl (g) → C₂H₄(g) + HCl(g)

Rate = k[C₂H₅Cl]

Here rate law is dependent on concentration of C₂H₅Cl. And power raised to C₂H₅Cl is 1. Hence order of reaction is 1.

To find dimensions of rate constant (K)- We know the rate of chemical reaction is measured in [concentration/time]. So, unit of K will be [Concentration]/[time][concentration]¹. i.e. [time]^-1

If concentration is in mol L^-1 and time is in sec. Then the dimensions of K will be sec^-1.

Ans (a) Rate = k[A][B]², Rate = 0 × 10^-6[0.1][0.2]²

Rate = 8 × 10^-9 mol L^-1sec^-1

(b) 2A +B = A₂B

=0.06+ 0.18 = 0.02

Situation when A remains 0.06 mol L^-1

Now, According to rate law, Rate = k[A][B]²

Rate = 2 × 10^-6[0.06][0.18]²

i.e. Rate = 3.888 × 10^-9 mol L^-1sec^-1

Initial rate of reaction is 8 × 10^-9 mol L^-1sec^-1. and rate when concentration of A is 0.06 mol L^-1 is 3.888 × 10^-9 mol L^-1sec^-1.

Ans 2NH₃(g) → N₂(g) + 3H₂(g)

Rate of zero order reaction is equal to rate constant. i.e. Rate = 2.5 × 10^-4mol L^-1sec^-1.

According to rate law,

-d[NH₃] / 2dt = d[N₂] / dt

2.5 × 10^-4mol L^-1sec^-1 = d[N₂] / dt

i.e. the rate of production of N₂ is 2.5 × 10^-4mol L^-1 sec^-1.

According to rate law,

-d[NH₃] / 2dt = d[H₂] / 3dt

d[H₂] / dt = -3 X d[NH₃] / 2dt

i.e. rate of formation of H₂ is 3 times rate of reaction = 3 × 2.5 × 10^-4mol L^-1sec^-1

= 7.5 × 10^-4mol L^-1sec^-1

Rate of formation of N₂ and H₂ is 2.5 × 10^-4 mol L^-1sec^-1 and 7.5 × 10^-4 mol L^-1sec^-1 respectively

Ans Rate of given chemical reaction will be represented as

-d[pCH₃OCH₃] / dt

Hence units of rate is bar min^-1

To find units of K, K = rate/[pCH₃OCH₃]^3/2

The unit of k = bar ^-1/2min^-1.

Units of Rate- bar min^-1. and Units of Rate constant K : bar ^-1/2min ^-1

Ans Factors affecting rate of reaction-

1) Temperature (increasing temperature increases rate of reaction)

2) Concentration or pressure of reactants. (Increasing concentration or pressure increases rate of reaction)

3) Presence or absence of a (Adding catalyst mostly increases the rate of reaction)

4) The surface area of solid (If reaction is processing over solid reactant then increasing its surface area increases rate of reaction)

5) Nature of reactants

Ans Let suppose reaction 

A ⇒B; having rate law, Rate = K [A]²

(i) If the concentration of A is doubled then the rate will affect by the square of concentration i.e. rate will become 2² = 4 times.

(ii) If the concentration of A is halved, then the rate will affect the square of the concentration i.e (1/2)² = 1/4

(i) Rate becomes 4 times of initial Rate (ii) Rate becomes 1/4 times of initial Rate

Ans Increase in temperature increases the rate constant of a reaction. as we know increase in temperature increases the rate of reaction to satisfy the equation Rate = k [concentration]ⁿ where n can be any real number. k have to increase as concentration is almost not changing over small temperature change.

Increasing the temperature by 10°c almost doubles the rate constant. This can be represented quantitatively by the help of the Arrhenius equation-

K = Ae^-Ea/RT, where k is rate constant, E_a is the activation energy, R is the universal gas constant, T is the absolute temperature.

Ans (i) Average rate of reaction over interval is [change in concentration]/[time taken] e.

[0.31 - 0.17] / [60-30] = 0.00467 mol L^-1 sec^-1

(ii) the pseudo first-order rate constant can be calculated by K = (2.303/t) log(C_i/C_t)

where K is Rate constant,

t is time taken,

C_i is initial concentration

C_t is Concentration at time t.

K = (2.303/30) log (0.55/0.31)

⇒K = 1.9 × 10^-2 sec^-1

(i) Average rate between 30 to 60 sec is 0.00467 mol L^-1sec^-1

(ii) Pseudo first order rate constant is 1.× 10^-2sec^-1

Q 4.9 A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled? 

Ans (i) Order is power raised to reactant in rate law, hence,

Rate = k[A][B]²

(ii) When the concentration of B is increased three times then the rate is affected by the square of The rate is increased by 9 Times.

(iii) When the concentration of reactant both A and B is doubled then the rate will have affected as square of reactant B and Two times of Reactant Overall increase in rate is 8 times

(a) When the concentration of B is increased by three times, the rate is increased by nine times

(b) When of both reactants is doubled, the Rate increases 8 times.

Ans When concentration of B is changed then rate of reaction doesn't change that means order with respect to B is 0. But when the concentration of A is doubled rate increased by 2.82 times i.e.2^1.5 = 2.82.Hence order with respect to A is 1.5.

Order with respect to A and B is 1.5 and 0, respectively.

Q 4.11 The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D 

Determine the rate law and the rate constant for the reaction.

Ans By comparing Experiment I and IV if we increase the concentration of A by 4 times then Rate also increased by 4 times. That means order with respect to A is 1.

By comparing Experiment II and III if we double the concentration of B Rate increases by 4 times that means order with respect to B is 2.

Rate law of reaction will be, Rate = k [A][B]²

To find K, K = rate/[A][B]2 i.e. K = 6.0 × 10^-3/[0.1][0.1]²

K = 6 mol^-2L²sec^-1

Order with respect to A and B is 1 and 2 respectively. And value of K(rate constant) is = 6 mol^-2L²sec-1

Ans As reaction is first order with respect to A and zero Order with respect to B. Then changing the concentration of B won’t affect the rate of reaction and increasing concentration of A ‘n’ times will increase the rate by ‘n’ times. By this logic lets fill the table- In first blank space concentration of A will be 0.2 mol L^-1 because the rate is doubled. In second blank space, Rate will be 8 × 10^-2mol L^-1min^-1 because the concentration of A is increased 4 Times. In third blank space concentration of A will be 0.1 mol L^-1 because the rate is same as in experiment I.

Ans Half life of first order reaction is, t_1/2 = ln2/K where t_1/2 is half life of first order reaction, K is rate constant of First order reaction.

(i) t_1/2 = ln2/200 s^-1

⇒t_1/2 = 0.693/200 s^-1 (∵ln2 = 0.693)

⇒t_1/2 = 0.003465 sec.

(ii) t_1/2 = ln2/2 min^-1

⇒t_1/2 = 0.693/2 min^-1 (∵ln2 = 0.693)

⇒t_1/2 = 0.3465 min

(iii) t_1/2 = ln2/4 year^-1

⇒t_1/2 = 0.693/4 years ^-1 (∵ln2 = 0.693)

⇒t_1/2 = 0.17325 year.

Half life of 3 reactions are 0.003465 sec, 0.3465 min, 0.17325 year, respectively.

Ans Radio active decay occurs via first order rate law,

t_1/2 = 5730 years. rate constant(k) of given decay is 0.693/t_1/2

⇒0.693/5730 = 1.2 × 10^-4 year^-1

By first order integrated rate law age of the sample will be,

T  = ( 2.303 / 1.2 X 10^-4 ) log (A₀/A_t)

where T is the age of the sample, A₀ is the initial activity of the sample. and A_t is the activity of the sample at any time t

T  = ( 2.303 / 1.2 X 10^-4 ) log (A₀/0.8 A₀)

T = 0.18 × 10⁴ years.

Age of given sample is 0.18 × 10⁴ years.

Q 4.15 The experimental data for decomposition of N₂O₅ [2N₂O₅ → 4NO₂ + O₂ ] in gas phase at 318K are given below:

(i) Plot [N2O5 ] against t.

(ii) Find the half-life period for the reaction.

(iii) Draw a graph between log[N2O5 ] and t.

(iv) What is the rate law ?

(v) Calculate the rate constant.

(vi) Calculate the half-life period from k and compare it with (ii).

Ans

(iv) As log [N₂O₅] vs time is a straight line given reaction is first Hence its rate law will be, Rate = k[N₂O₅]

(v) The slope of above graph is slope = 0.000209 K = 303 × slope

⇒4.82 × 10^-4sec^-1

Now, t_1/2 = 0.693/K.

⇒0.693/4.82 × 10^-4

⇒t_1/2 = 1438 sec. which is almost equal to (ii)

Ans Given:

Order of the reaction = 1

Let, Initial concentration [R]^°= x

Final concentration [R] = x/16

Rate constant k = 60 s^-1

We know, time

t= 2.303 / k log R₀ / R

t = 2.303 / 60 log (x/x/16)

t = 2.303 / 60s^-1 log (1/1/16)

t = 2.303 X log 16 / 60s^-1

Solving, we get t = 4.6 × 10^-2s

Ans Initial concentration, [R]_° = 1μg

Final concentration, [R] Half-life t_1/2 = 28.1 yrs Solution:

We know, t_1/2 = 0.693/k Where, k – rate constant

⇒k = 0.693/ t_1/2

⇒k = 0.693/(28.1 yrs)

⇒k = 0.0246 yrs^-1

Also, t = 2.303 / k log R₀ / R

If t = 10yrs, then, using the formula, we get,

t = 2.303 / k log R₀ / R

10 = 2.303 / 0.0246 log 1 / R₁₀ 

log 1 / R₁₀  = 0.0246 X 10 / 2.303

log 1 / R₁₀ = 0.246 / 2.303

1 / R₁₀ = antilog (0.246 / 2.303)

1 / R₁₀ = 1.278                                                                                                                                                                                        

⇒  R₁₀ = 0.7824μg

If t = 60yrs, then again, we get,

60 = 2.303 / 0.0246 log 1 / R₆₀ 

log 1 / R₆₀  = 0.0246 X 60 / 2.303

log 1 / R₆₀ = antilog (1.476 / 2.303)

1 / R₆₀ = 4.374                                                                                                                                                                             

⇒[R]₆₀ = 0.228 μg

∴0.2278μg of ⁹⁰Sr will be left after 60 years.

Ans Let, initial concentration be [R]_°

Concentration at 90% completion be ((100-90)/100)×[R]_°

∴Concentration at 90% be 0.1[R]_°

Concentration at 99% completion be ((100-99)/100)× [R]_°∴Concentration at 99% be 0.01[R]_°

_{we know time, t= 2.303/K log R0 / R}

Time taken for 90% completion is 

T₉₀ = 2.303 / K log R₀ / 0.1 R₀ 

T₉₀ = 2.303 / K log 1 / 0.1

T₉₀ = 2.303 / K log 10 / 1

T₉₀ = 2.303 / K

Time taken for 99% completion is

T₉₉ = 2.303 / K log R₀ / 0.01 R₀ 

T₉₉ = 2.303 / K log 1 / 0.01

T₉₉ = 2.303 / K log 100 / 1

T₉₉ = 2 X 2.303 / K

T₉₉ = 2 T₉₀

Hence, the time taken to complete 9% of the first order reaction is twice the time required for the completion of 90% of the reaction.

Ans Given:

Time t = 40 min

When 30% decomposition is undergone, 70% is the concentration.

We know, time taken  

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]₀ - initial concentration

[R] - concentration at time 't'

40 = 2.303/K log R₀ / 0.7 R₀

40 = 2.303/K log 1 / 0.7

40 / 0.1549 = 2.303 / k

⇒258.23 = (2.303/k)

We know, Half-life t_1/2 = 0.693/k

Which can be written as, t_1/2 = 0.3010 × (2.303/k)

⇒t_1/2 = 0.3010 × 258.23

⇒t_1/2 = 77.72 min

Q 4.20 For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. 

Calculate the rate constant.

Ans When t = 0, the total partial pressure is P₀ = 35.0 mm of Hg

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀ Hence, P₀-p = P_t-2( P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]_° -Initial concentration of reactant

[R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

t= 2.303/K log P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

→ equation 1

At time t = 360 s, P_t = 54 mm of Hg and P₀ = 30 mm of Hg, Substituting in equation 1,

360 = 2.303/k log 30 / (2X30)-54

k= 2.175 X 10^-3 s^-1

At time t = 720 s, P_t = 63 mm of Hg and P₀ = 30 mm of Hg, Substituting in equation 1,

720 = 2.303 / k log 30 / (2X30) - 63

Thus, k = 2.235 × 10^-3 s^–1

Taking average, k = (2.235 × 10^-3 s^–1 + 2.175 × 10^-3 s^–1)/2

∴k = 2.21 × 10^-3 s^–1.

Q 4.21 The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume. SO₂Cl_2(g)→ SO_2(g) + Cl_2(g)

Calculate the rate of the reaction when total pressure is 0.65 atm. 

Ans When t = 0, the total partial pressure is P₀ = 0.5 atm

When time t = t, the total partial pressure is P_t = P₀ + p

P₀-p = P_t-2p, but by the above equation, we know p = P_t-P₀

Hence, P₀-p = P_t-2(P_t-P₀)

Thus, P₀-p = 2P₀ – P_t

We know that time

t= 2.303/K log R₀ / R

Where, k- rate constant

[R]_° -Initial concentration of reactant [R]-Concentration of reactant at time ‘t’

Here concentration can be replaced by the corresponding partial pressures.

Hence, the equation becomes,

t= 2.303/K log P₀ / P₀ - P

t= 2.303/K log P₀ / 2P₀ - P_t

→ equation 1

At time t = 100 s, P_t = 0.6 atm and P₀ = 0.5 atm,

Substituting in equation 1,

100 = 2.303/k log 0.5 / (2X0.5) - 0.6

Thus, k = 2.231 × 10^-3 s^-1

The rate of reaction R = k × P_S02Cl₂

When total pressure P_t = 0.65 atm and P₀ = 0.5 atm, then

[P_S02Cl₂ = 2P₀-P_t

Thus, substituting the values, P_S02Cl₂ = 2(0.5)-0.6 = 0.35 atm

R = k × P_S02Cl₂ = 2.231 × 10^-3 s^–1 × 0.35

Rate of the reaction R = 7.8 × 10^-4atm s^–1

Q 4.22 The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between ln k and 1/T and calculate the values of A and Ea . Predict the rate constant at 30° and 50°C. 

Ans To convert the temperature in °C to °K we add 273 K.

The graph is given as:

The Arrhenius equation is given by k = Ae^-Ea/RT

Where, k- Rate constant

A- Constant

E_a-Activation Energy

R- Gas constant

T-Temperature

Taking natural log on both sides,

ln k = ln A-(E_a/RT)...................... equation 1

By plotting a graph, ln K Vs 1/T, we get y-intercept as ln A and Slope is –E_a/R.

Slope = (y₂-y₁)/(x₂-x₁)

By substituting the values, slope = -12.301

⇒–E_a/R = -12.301

But, R = 8.314 JK^-1mol^-1

⇒ _aE= 8.314 JK^-1mol^-1 × 12.301 K

⇒ _aE= 102.27 kJ mol^-1

Substituting the values in equation 1 for data at T = 273K

(∵ At T = 273K, ln k =-7.147)

On solving, we get ln A = 37.911

∴ A = 2.91×10⁶

When T = 30⁰C, hence T = 30 + 273 = 303K

⇒ ln k =-2.8

⇒ k = 6.08x10^-2s^-1.

When T = 50⁰C, hence T = 50 + 273 = 323K

Substituting in equation 1,

A = 2.91 × 10⁶, E_a = 102.27 kJ mol^-1

Ln K = ln(2.91 × 10⁶)- 102.27/(8.314 × 323)

Thus, ln k = -0.5 and k = 0.607s^-1.

Ans Given,

k = 2.418 × 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 × 10³J mol^-1

The Arrhenius equation is given by k = Ae^-Ea/RT Taking natural log on both sides,

Ln k = ln A-(E_a/RT) Substituting the values,

ln(2.418 × 10^-5 ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10¹² s^-1(approximately)

Ans Given,

k = 2.0 × 10^–2s^-1

time t = 100s

Concentration [A₀] = 1.0 mol L^-1

We know,

On substituting the values,  Log(1/[A]) = 2.303/2

Log[A] = -2.303/2 [A] = 0.135 mol L^–1

Ans t_1/2 = 3.00 hours

We know, t_1/2 = 0.693/k

∴k = 0.693/3 k = 0.231 hrs^-1

We know, time  

Where, k- rate constant

[R]_° -Initial concentration

[R]-Concentration at time ‘t’

Thus, substituting the values,

log([R]₀/[R]) = 0.8

log([R]/[R]₀) = -0.8

[R]/[R]₀ = 0.158

Hence, 0.158 fraction of sucrose remains.

Q 4.26 The decomposition of hydrocarbon follows the equation

k = (4.5 × 10¹¹s^–1) e^-28000K/T. Calculate E_a

Ans

The given equation is

k = (4.5 x 10¹¹ s^-1) e^-28000K/T (i)

Comparing, Arrhenius equation

k = Ae ^-E/RT (ii)

We get, E_a / RT = 28000K / T

⇒E_a = R x 28000K

= 8.314 J K^-1mol^-1 × 28000 K

= 232792 J mol^–1 or 232.792 kJ mol^–1

Ans

We know, The Arrhenius equation is given by k = Ae^-Ea/RT Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Thus, log k = log A -(E_a/2.303RT)......................................... eqn 1

The given equation is log k = 14.34 – 1.25 × 10⁴K/T.............. eqn 2

Comparing 2 equations, E_a/2.303R = 1.25 × 10⁴K

E_a = 1.25 × 10⁴K × 2.303 × 8.314

E_a = 239339.3 J mol^-1 (approximately) E_a = 239.34 kJ mol^-1

Also, when t_1/2 = 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 × 10^-3 min^-1 k = 4.51 × 10^-5s^–1

Substitute k = 4.51 × 10^-5s^–1 in eqn 2,

log 4.51 × 10^-5 s^–1 = 14.34 – 1.25 × 10⁴K/T

log(0.654-5) = 14.34– 1.25 × 10⁴K/T T = 1.25 × 10⁴/[ 14.34- log(0.654-5)] T = 668.9K or T = 669 K

Ans From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s ^-1

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s ^-1

E_a = 60 kJ mol ^-1 = 6.0 × 10⁴ J mol ^-1

Then,

→ 0.5229 = 3133.627 × (T₂-283)/(283 × T₂)

→ 0.0472T₂ = T₂-283 T₂ = 297K or T₂ = 24⁰ C

Ans We know, time t = (2.303/k) × log([R]₀/[R])

Where, k- rate constant

[R]₀-Initial concentration

[R]-Concentration at time ‘t’

At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) × log ([R]₀/0.9[R]₀)

t = (2.303/k) × log (1/0.9) t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining t’ = (2.303/k’) × log ([R]₀/0.75[R]₀)

t’ = (2.303/k’) × log (1/0.75) t’ = 2.2877 / k'

But, t = t’

0.1054 / k = 2.2877 / k' k' / k = 2.7296

From Arrhenius equation, we obtain log k₂/k₁ = (E_a / 2.303 R) × (T₂ - T₁) / T₁T₂

Substituting the values,

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1 We know, log k = log A –E_a/RT

Log k = log(4 × 10¹⁰)-(76.64kJ mol^-1/(8.314 × 318) Log k = -1.986

∴k = 1.034 x 10^-2 s ^-1

Ans

Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know that from the Arrhenius equation, we obtain

On solving, we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1