Question

The rate constant for the decomposition of \(N_2O_5\) at various temperatures is given below:

T/°C	0	20	40	60	80
105 x k/s-1	0.0787	1.70	25.7	178	2140

Draw a graph between ln k and \(\frac 1T\) and calculate the values of \(A\) and \(E_a\).
Predict the rate constant at 30 ºC and 50 ºC.

Answer 1

From the given data, we obtain

T/°C	0	20	40	60	80
T/K	273	293	313	333	353
\(\frac 1T\)/K-1	3.66×10-3	3.41×10-3	3.19×10-3	3.0×10-3	2.83 ×10-3
105 x k/s-1	0.0787	1.70	25.7	178	2140
ln k	- 7.147	- 4.075	- 1.359	- 0.577	3.063

Slop of the line,

\(\frac {y_2-y_1}{x_2-x_1} = -12.301\ K\)

According to Arrhenius equation,

\(Slope = -\frac {E_a}{R}\)
\(E_a = - Slope \times R\)
E_a =-(-12.301 k) x (8.314 JK^-1mol^-1)
E_a =102.27 kJ mol^-1
Again,

\(ln\  k = ln \ A-\frac {E_a}{RT}\)

\(ln\  A = ln \ k-\frac {E_a}{RT}\)

\(When\  T = 273\  k\)
\(ln \ k = -7.147\)
Then, 

\(ln\  A = - 7.17 + \frac {102.27 \times 10^3}{8.314 \times 2.73}\)

= \(37.911\)
Therefore, \(A = 2.91 \times 10^6\)

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when T = 30+273 K = 303 K
\(\frac 1T\) = 0.0033 K = 3.3 x 10^-3 K
Then, 
At \(\frac 1T\) = 3.3 x 10^-3 K
ln k = - 2.8
Therefore, \(k = 6.08 \times  10^{-2} s^{-1}\)
Again when T = 50 + 273 = 323 K
\(\frac 1T\) = 0.0031 K = 3.1 x 10^-3 K
Then,
At \(\frac 1T\) = 3.1 x 10^-3 K
ln k = - 0.5
Therefore, \(k = 0.607 \ s^{-1}\)