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Chemical Kinetics Class 12th

4.13 The decomposition of dimethyl ether leads to the formation of CH₄ , H₂ and CO and the reaction rate is given by Rate = k [CH₃OCH₃ ] 3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate= k(PCH₃OCH₃)^3/2 If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

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Payal Gupta

Contributor-Level 10

4.13 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

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Chemical Kinetics Class 12th

4.12 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 * 10^–4 mol^–1 L s^–1?

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Payal Gupta

Contributor-Level 10

4.12 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

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Chemical Kinetics Class 12th

4.11 For the reaction 2A + B → A₂B the rate = k[A][B]₂with k = 2.0 * 10^–6 mol^–2 L2 s^–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^–1, [B] = 0.2 mol L^–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^–1.

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Payal Gupta

Contributor-Level 10

4.11 (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.

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Chemical Kinetics Class 12th

4.10 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]

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Payal Gupta

Contributor-Level 10

4.10 (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 ra

...more

4.10 (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.

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Chemical Kinetics Class 12th

4.9 The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

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Payal Gupta

Contributor-Level 10

4.9 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/RTrepresents 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

...more

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Chemical Kinetics Class 12th

4.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea .

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Payal Gupta

Contributor-Level 10

4.8 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,

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Chemical Kinetics Class 12th

4.7 What will be the effect of temperature on rate constant?

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Payal Gupta

Contributor-Level 10

4.7 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.

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Chemical Kinetics Class 12th

4.6 Time required to decompose SO2Cl2 to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

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Payal Gupta

Contributor-Level 10

4.6 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

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Chemical Kinetics Class 12th

4.5 A first order reaction has a rate constant 1.15 * 10-3 s-1. How long will 5 g of this reactant take to reduce to 3 g?

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Payal Gupta

Contributor-Level 10

4.5 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

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Chemical Kinetics Class 12th

4.4 The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

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Payal Gupta

Contributor-Level 10

4.4 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

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