# please solve the following question:- A reaction SO2Cl2 ---->SO2 + Cl2 is a first order reaction with half life time period 3.5*104s at 320C . What percentage of SO2Cl2 would be decomposed on heating at 320 C FOR 90 mins.

Solution-

Total time T = n x t1/2

t1/2 = 3.5 x 104 sec =35000 seconds

T = 90 minutes = 5400 seconds

Let the initial amount is N° and the final amount of reactant is N

$\mathrm{K}=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{\mathrm{N}}^{\mathrm{o}}}{\mathrm{N}}$  ------(1)

K = 0.693/ t1/2  ------(2
)

Therefore from equation (2)

K = 1.98 x 10-5

Putting the value of k and t in equation (1)

log No/N = 0.0000198 $×$ 5400 = 0.1069

Taking antilog on both sides,

N°/N = antilog 0.1069

N°/N = 1.279

Therefore remaining amount (N) = 0.7818 N°

% Amount decomposed = (N° - 0.7818 N°) x 100/N°

= 21.81%

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