Question

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

Experiment	Time/$s^{-1}$	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)

$S{O}_{2}C{l}_2$ $\to$ $S{O}_2+C{l}_2$
Initial pressure $p_0$ $0$ $0$
Pressure at time $t$ $p_0-p$ $p$ $p$
Let initial pressure, $P_0$ $\propto$ $R_0$
Pressure at time $t$, $p_t$ $=p_0-p+p+p$ $=p_0+p$
Pressure of reactants at time $t$, $p_0-p$ $=2p_0-p_t\propto R$
$k=\frac{2.303}{t}log$ $\frac{p_0}{2p_0-p_t}$
$=\frac{2.303}{100}log$ $\frac{0.5}{2\times 0.5-0.6}$ $=\frac{2.303}{100}log1.25$
$=2.2318\times 10^{-3}$ $s^{-1}$
Pressure of $S{O}_{2}C{l}_2$ at time $t$ $\left(p_{s{o}_{2}c{l}_2}\right)$
$=2p_0-p_t$ $=2\times 0.50$ $-0.65atm$ $=0.35atm$
Rate at 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 $s^{-1}$