Question

Observation no.	Time (in min)	$P_x\left(inmmHg\right)$
1	0	800
2	100	400
3	200	200

At constant temperature and volume, $X$ decomposes according to first order kinetics as follows:
$2X\left(g\right)\to 3Y\left(g\right)+2Z\left(g\right)$
where $P_x$ is the partial pressure of $X$.

Which of the following statements is true for the given reaction?

• A. Rate constant of reaction is $6.93\times {10}^{-3}{min}^{-1}$
• B. At $200min$, 75% of the reaction completes
• C. When $P_x=700mmHg$, the total pressure of the system is $950mmHg$
• D. All of the above

Answer 1

The correct option is D All of the above

$P_x$ is the partial pressure of $X$
$P_0-P=P_x\text{at time 't'}$
where,
$P_0$ is initial pressure at $t=0$
$P$ is pressure of reactant which is already converted to product at time 't'.

Applying intergrated rate law of first order kinetics,

$k=\frac{1}{t}ln\left(\frac{P_0}{P_0-P}\right)\text{(In terms of pressure})$

$k=\frac{2.303}{t}\log \frac{P_o}{P_x}$

$P_0=800mmHg$
At time, $t=100min$
$P_x=400mmHg$

$k_1=\frac{2.303}{100}\log \left(\frac{800}{400}\right)$
$k_1=\frac{2.303}{100}\log 2=6.932\times {10}^{-3}{min}^{-1}$
Similary, at time, $t=200min$
$k_2=\frac{2.303}{200}\log \left(\frac{800}{200}\right)$
$k_2=\frac{2.303}{200}\log 4=6.932\times {10}^{-3}{min}^{-1}$
Since k value is same in both, hence verified that it must be following the first order kinetics.

$\text{Rate constant,}k=6.932\times {10}^{-3}{min}^{-1}$
(C)
At 75% completion of reaction

$k=\frac{2.303}{t}\log \frac{100}{25}$

$6.932\times {10}^{-3}=\frac{2.303}{t}\log \frac{100}{25}$
$6.932\times {10}^{-3}=\frac{2.303}{t}\log 4$

$6.932\times {10}^{-3}=\frac{2.303}{t}\times 0.6$
$t=200min$

(D)
$2X\left(g\right)\to 3Y\left(g\right)+2Z\left(g\right)$
$t=0:80000$
$t=t:(800-2P)3P2P$
When the partial pressure of X is $700mmHg$
then,
$800-2P=700$
$2P=100$
$P=50mmHg$
$\text{Total pressure}=800-2P+3P+2P$
$=800+3P$
$=800+3\times 50$
$=950mmHg$