Ncert Exercises & Solutions

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.

S O_{2} C l_{2 (g)} \to S O_{2 (g)} + C l_{2 (g)}

A t l = 0 P_{0} 0 0

A t t = t P_{0} - p p p

P_{i} = (P_{0} p) + p + p

After time, t, total pressure,

P_{1} = P_{0} + p

\Rightarrow p = P_{t} - P_{0}

P_{0} - p = P_{0} - (P_{t} - P_{0})

T h e r e f o r e,

= 2 P_{0} - P_{t}

For a first order reaction

k = \frac{2.303}{t} \log \frac{P_{0}}{P_{0} - p}

- \frac{2.303}{t} \log \frac{P_{0}}{2 P_{0} - P_{t}}

k = \frac{2.303}{100 s} \log \frac{0.5}{2 \times 0.5 - 0.6 t =} 100 s,

w h e n

= 2.231 \times 10^{- 3} s^{- 1}

When

P_{t} =

0.65 atm,

P^{0} + p = 0.65

\Rightarrow p = 0.65 - P_{0}

= 0.65 - 0.5

= 0.15 a t m

Therefore, when the total pressure is 0.65 atm, pressure of SOCl

_{2}

P_{s o c l_{2}} = P_{0} - P

= 0.5 - 0.15

= 0.35 a t m

Therefore, the rate of equation, when total pressure is 0.65 atm, is given by,

Rate = k

(P_{s o c l_{2}})

= (2.23 \times 10^{- 3} s^{- 1} (0.35 a t m)

= 7.8 \times 10^{- 4} a t m s^{- 1}