Question

On heating, arsine $\left(As{H}_3\right)$ decomposes according to first order kinetics as follows:
$2As{H}_3\left(g\right)\to 2As\left(s\right)+3H_2\left(g\right)$
The total pressure measured at constant temperature and constant volume varies with time as given:

$t\left(min\right)$	0	5	7.5	10
$P_t\left(mmHg\right)$	760	836	866	897

Calculate the rate constant from above data
Take:
$log\left(1.25\right)=0.096log(1.38=0.142log(1.56)=0.194$

• A. $9.8mi{n}^{-1}$
• B. $0.98mi{n}^{-1}$
• C. $4.4mi{n}^{-1}$
• D. $0.044mi{n}^{-1}$

Answer 1

The correct option is D $0.044mi{n}^{-1}$

For first order, the rate equation is

$k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}=\frac{2.303}{t}log\frac{P_0}{P}$
$P_0$ is initial pressure
$P$ is pressure of reactant at time 't'

Given,
$P_0=760mmHg$

The decomposition reaction is:
$2As{H}_3\left(g\right)\to 2As\left(s\right)+3H_2\left(g\right)$
$Att=0:P_0-0$
$Att=t:P_0-2x-3x$
$\text{Total pressure,}{P}_t=P_0-2x+3x=P_0+x$
$x=P_t-P_0$
$P_{As{H}_3}=(P_0-2x)=P_0-2P_t+2P_0=3P_0-2P_t$

After 5 minutes,
$P_{As{H}_3}=(3\times 760)-(2\times 836)$
$P_{As{H}_3}=608mmHg$
$k=\frac{2.303}{5}lo{g}_{10}\frac{760}{608}=0.0446mi{n}^{-1}$

After 7.5 minutes,
$P_{As{H}_3}=(3\times 760)-(2\times 866)$
$P_{As{H}_3}=548mmHg$
$k=\frac{2.303}{7.5}lo{g}_{10}\frac{760}{548}=0.0436mi{n}^{-1}$

After 10 minutes, $P_{As{H}_3}=(3\times 760)-(2\times 897)$
$P_{As{H}_3}=486mmHg$
$k=\frac{2.303}{10}lo{g}_{10}\frac{760}{486}=0.0447mi{n}^{-1}$

The constancy of k shows that, the reaction follows first order kinetics.