Question

For a $I^{st}$ order decomposition of $S{O}_{2}C{l}_2\left(g\right)$, to $S{O}_2\left(g\right)$ and $C{l}_2\left(g\right)$, an experiment is carried out and the following graph is plotted:

$S{O}_{2}C{l}_2\left(g\right)\to S{O}_2\left(g\right)+C{l}_2\left(g\right)$


What is the rate of the reaction at t = 10 min in $\left(mol{L}^{-1}se{c}^{-1}\right)$

• A. $0.2\times {10}^{-3}$
• B. $4.6\times {10}^{-4}$
• C. $7.7\times {10}^{-6}$
• D. $1.15\times {10}^{-5}$

Answer 1

The correct option is C $7.7\times {10}^{-6}$

$S{O}_{2}C{l}_2\left(g\right)\to S{O}_2\left(g\right)+C{l}_2\left(g\right)$

t = 0 a
t = t (a -x ) x x

rate = $k\times \left[S{O}_{2}C{l}_2\right]$

at t = 10 min

from graph

$log(a-x)$ = -3

(a - x) = ${10}^{-3}mol{L}^{-1}$

rate = $k\times \left[S{O}_{2}C{l}_2\right]$
= $7.7\times {10}^{-3}se{c}^{-1}\times {10}^{-3}mol{L}^{-1}$
= $7.7\times {10}^{-6}mol{L}^{-1}se{c}^{-1}$