Question

The following data were obtained during the first order thermal decomposition of $S{O}_{2}C{l}_2$ at a constant volume.
$S{O}_{2}C{l}_{2\left(g\right)}\to S{O}_{2\left(g\right)}+C{l}_{2\left(g\right)}$

Experiment	Time / $s$	Total pressure/atm
1	0	0.5
2	100	0.6

What is the rate of reaction when total pressure is 0.65 atm?

• A. $0.35$ atm $s^{-1}$
• B. $2.235\times 10^{-3}$ atm $s^{-1}$
• C. $7.8\times 10^{-4}$ atm $s^{-1}$
• D. $1.55\times 10^{-4}$ atm $s^{-1}$

Answer 1

Answer: (C)

$S{O}_{2}C{l}_{2\left(g\right)}\to S{O}_{2\left(g\right)}+C{l}_{2\left(g\right)}$
$k=\frac{2.303}{t}\log \frac{p_0}{2p_0-P_t}$
$=\frac{2.303}{100}\log \frac{0.5}{2\times 0.5-0.6}=\frac{2.303}{100}log1.25$
$=2.2318\times 10^{-3}{s}^{-1}$
Pressure of $S{O}_{2}C{l}_2$ at time t $\left(p_{S{O}_{2}C{l}_2}\right)$
$=2p_0-P_t=2\times 0.50-0.65$ atm
$=0.35$ atm
Rate of that time $=k\times p_{S{O}_{2}C{l}_2}$
$=\left(2.2318\times 10^{-3}\right)\times \left(0.35\right)=7.8\times 10^{-4}$ atm $s^{-1}$