A homogeneous ideal gaseous reaction \(\mathrm{AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}}\) is carried out in a 25 litre flask at \(27^\circ\mathrm{C}\). The initial amount of \(\mathrm{AB_2}\) was 1 mole and the equilibrium pressure was 1.9 atm. The value of \(\mathrm{K_p}\) is x \(\times\) \(10^{-2}\). The value of x is _________. (Integer answer)
[R = 0.08206 \(\mathrm{dm^3\ atm\ K^{-1}\ mol^{-1}}\)]
[R = 0.08206 \(\mathrm{dm^3\ atm\ K^{-1}\ mol^{-1}}\)]
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Correct Answer: 73
Solution and Explanation
\(\mathrm{AB_2 \rightleftharpoons A + 2B}\): moles \(1-\alpha,\ \alpha,\ 2\alpha\); \(n_t = 1 + 2\alpha = 1.9295 \Rightarrow \alpha = 0.4647\).
Moles: \(\mathrm{AB_2}\) = 0.5353, A = 0.4647, B = 0.9295. Partial pressures: \(p = P \times n/n_t\): \(p_{AB_2} = 0.527\), \(p_A = 0.4577\), \(p_B = 0.9153\) atm.
\(K_p = \dfrac{p_A\, p_B^2}{p_{AB_2}} = \dfrac{0.4577 \times 0.8378}{0.527} = 0.727 \approx 72.7 \times 10^{-2}\), so x is about 73.
(NTA accepted the range 72 to 75.)
Answer: 73
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