JEE MAIN - Chemistry (2021 - 22th July Evening Shift - No. 19)
Value of KP for the equilibrium reaction
N2O4(g) $$\rightleftharpoons$$ 2NO2(g) at 288 K is 47.9. The KC for this reaction at same temperature is ____________. (Nearest integer)
(R = 0.083 L bar K$$-$$1 mol$$-$$1)
N2O4(g) $$\rightleftharpoons$$ 2NO2(g) at 288 K is 47.9. The KC for this reaction at same temperature is ____________. (Nearest integer)
(R = 0.083 L bar K$$-$$1 mol$$-$$1)
Answer
2
Explanation
For the equilibrium reaction
$ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) $
the relationship between $ K_P $ and $ K_C $ is given by the equation:
$ K_P = K_C (RT)^{\Delta n} $
where $ \Delta n $ is the change in the number of moles of gas, $ R $ is the ideal gas constant, and $ T $ is the temperature in Kelvin.
For the given reaction:
$ \Delta n = 2 - 1 = 1 $
Given:
$ K_P = 47.9 $
$ R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} $
$ T = 288 \, \text{K} $
Substitute these values into the equation:
$ 47.9 = K_C (0.083 \times 288)^{1} $
$ K_C = \frac{47.9}{0.083 \times 288} $
Calculate:
$ 0.083 \times 288 = 23.904 $
$ K_C = \frac{47.9}{23.904} $
$ K_C \approx 2.0033 $
Rounding to the nearest integer, the value of $ K_C $ is
$ \boxed{2} $
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