Question

The rate constant for the decomposition of $N_2O_5$ at various temperatures is given below:

**T/°C **	0	20	40	60	80
10 5 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
10 5 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,

$Slope = -\frac {E_a}{R}$
$E_a = - Slope \times R$
Ea =-(-12.301 k) x (8.314 JK-1mol-1)
Ea =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, $k = 0.607 \ s^{-1}$