Question: 73. For the reaction \mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g}), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K . [Given: \mathrm{R}=0.0831 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ] \mathrm{K}_{\mathrm{P}} for the reaction at 1000 K is
(1) 83.1
(2) 2.077 \times 10^{5}
(3) 0.033
(4) 0.021
Answer: Option (3)
Explanation:
For an elementary reversible reaction, the equilibrium constant in terms of concentration is given by the ratio of rate constants as \mathrm{K}_{\mathrm{c}}=\frac{k_f}{k_b}.
Given that the backward rate constant is higher than the forward rate constant by a factor of 2500,
we have \frac{k_b}{k_f}=2500.
Therefore, \mathrm{K}_{\mathrm{c}}=\frac{k_f}{k_b}=\frac{1}{2500}=0.0004.
For the reaction \mathrm{A}(\mathrm{g}) \rightleftharpoons 2\mathrm{B}(\mathrm{g}),
the change in number of moles of gaseous species is \Delta n = 2-1 = 1.
The relation between \mathrm{K}_{\mathrm{P}} and \mathrm{K}_{\mathrm{c}} is \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta n}.
At 1000 K, \mathrm{RT}=0.0831 \times 1000 = 83.1.
Thus, \mathrm{K}_{\mathrm{P}}=0.0004 \times 83.1 = 0.033 approximately.
Hence, the correct answer is Option (3).