Chemical EquilibriumMedium
Question
For the reaction $A(g) \rightleftharpoons 2\text{ }B(\text{ }g)$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K .[Given : $R = 0.0831\text{ }L\text{ }atm{\text{ }mol}^{- 1}{\text{ }K}^{- 1}$ ]
$K_{P}$ for the reaction at 1000 K is
Options
A.83.1
B.$\ 2.077 \times 10^{5}$
C.0.033
D.0.021
Solution
Sol. [Key Concept: Kₚ from forward/backward rate constants (K = k_f/k_b; Kₚ = K_c(RT)^Δn)] $K_{C} = \frac{k_{f}}{k_{b}} = \frac{1}{2500}$
$$\begin{matrix} & K_{P} = K_{C}(RT)^{\Delta n_{g}}\ \left( \Delta n_{g} = 2 - 1 = 1 \right) \\ & \ = \frac{1}{2500} \times 0.0831 \times 1000 \\ & \ = 0.033 \end{matrix}$$
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