Question

The first-order reaction, so2cl2 → so2 + cl2, has a rate constant equal to 2.20 × 10-5 s-1 at 593 k. what percentage of the initial amount of so2cl2 will remain after 2.00 hours?

Answer 1

For the 1st order reactions,rate constant (k) is mathematically expressed as

k = $\frac{2.303}{t}log \frac{Co}{Ct}$
where, t = time
Co = initial conc. of reactant
Ct = conc. of reactant after time 't'

Given: k = 2.20 × 10^-5 s-1, t = 2 hours = 7200 s

Therefore, we have
2.20 × 10^-5 = $\frac{2.303}{7200}log \frac{100}{Ct}$
∴ $log\frac{100}{Ct}$ = 0.06877
∴, $\frac{100}{Ct}$ = 1.1716
∴, Ct = 85.35%

Thus, 85.35 % of the initial amount of SO2Cl2 will remain after 2.00 hours.