Chemical Kinetics

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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 * 1

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

4.9 Given-

Energy of activation, Ea= 209.5 kJ mol–1

Ea 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

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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 ex

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New answer posted•a year ago
S
a year ago

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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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

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 r

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New answer posted•a year ago
S
a year ago

4.3 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

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