Question

The following data were obtained during the first order thermal composition 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	$Time/s^{-1}$	Total pressure (atm)
$1$ $2$	$0$ $100$	$0.5$ $0.6$

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

• A. $0.35atm{s}^{-1}$
• 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

The correct option is C $7.8\times {10}^{-4}atm{s}^{-1}$

$\underset{p_o{p}_o-p}{SOC{l}_2}\to \underset{0p}{S{O}_2}+\underset{0att=0patt=t}{C{l}_2}$

Pressure at time $t,P_t=p_0-p+p+p=p_0+p$

Pressure of reactants at time $t,p_0-p=2p_0-p_t\propto R$

$k=\frac{2.303}{t}\log \frac{p_0}{2p_0-P_t}$

$=\frac{2.303}{100}\log \frac{0.5}{2\times 0.5-0.6}=\frac{2.303}{100}\log 1.25$

$=2.2318\times {10}^{-3}{s}^{-1}$

Pressure of $S{O}_{2}C{l}_2$ at time $t\left(p_{S{O}_{2}C{l}_2}\right)$

$=2p_0-P_t=2\times 0.50-0.65atm=0.35atm$

Rate at that time $=k\times p_{S{O}_{2}C{l}_2}$

$=(2.2318\times {10}^{-3})\times \left(0.35\right)=7.8\times {10}^{-4}atm{s}^{-1}$.