Question

A constant temperature and volume X decomposes as,
$2X(g) \to 3Y(g) + 2Z(g)$
${P_X}$ is the partial pressure of X.

Observation No.	Time (min)	${P_X}$ (in mm of Hg)
1	0	800
2	100	400
3	200	200

(i) What is the order of the reaction with respect to X?
(ii) Find the rate constant.
(iii) Find the time for \[75\% \] completion of the reaction.
(iv) Find the total pressure when pressure of X is \[700{\text{ }}mm{\text{ }}of{\text{ }}Hg\] .

Answer 1

Hint: The various rate laws in the chemical kinetics have different relations with respect to the reactants and their concentration. The coefficients of the reactants become the power in the rate law and addition of the total power gives the order of the reaction. Based on the knowledge of this order of the reaction, we apply various relations to find the rate constant, time for completion of a certain amount of reaction, total pressure when the value of partial pressure is provided.

Complete answer:
(i) According to the rate law of the above reaction,
$R = k{\left[ X \right]^2}$
Thus, the order of the reaction with respect to X is 2.
(ii) Let us apply the first order of reaction kinetics or what we say, the integrated rate law of first order kinetics, which states that:
${k_1} = \dfrac{{2.303}}{t}\log \dfrac{{{P_o}}}{{{P_t}}}$ ….. (i)
Taking observation 1 and 2 and substituting the values in equation (i), we have:
$
  {k_1} = \dfrac{{2.303}}{{100}}\log \dfrac{{800}}{{400}} \\
  {k_1} = 6.932 \times {10^{ - 3}}{\min ^{ - 1}} \\
$
Taking observation 2 and 3 and substituting the values in equation (i), we have:
$
  {k_1} = \dfrac{{2.303}}{{100}}\log \dfrac{{400}}{{200}} \\
  {k_1} = 6.932 \times {10^{ - 3}}{\min ^{ - 1}} \\
$
Thus, in both the reactions, the value of rate constant is the same. This proves that the reaction is of first order.
Thus, the rate constant is equal to $6.932 \times {10^{ - 3}}\min {}^{ - 1}$ .
(iii) In order to find out the time required for the \[75\% \] completion of the reaction, we use the equation:
${k_1} = \dfrac{{2.303}}{t}\log \dfrac{{{P_o}}}{{{P_t}}}$
Substituting the values and solving, we have:
$\begin{gathered}
  6.932 \times {10^{ - 3}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{25}} \\
  {t_{75\% }} = 200\min \\
\end{gathered} $
(iv) $2X(g) \to 3Y(g) + 2Z(g)$
Initial: 800 0 0
Final: (800-2P) 3P 2P
When the pressure of X becomes \[700{\text{ }}mm{\text{ }}of{\text{ }}Hg\],
$
 800 - 2P = 700 \\
 P = 50 \\
$ mm of Hg
Hence, total pressure = $800 - 2P + 3P + 2P = 800 + 3(50) = 950$ mm of Hg

Note:
The order of a reaction is identified by comparing the various values of the different parameters given in the question whereas the order of a particular reactant is identified by its stoichiometric coefficient in the reaction.