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Standard XII

Chemistry

Stoichiometric Calculations

Two moles of ...

Question

Two moles of gas $A_{2}$ are mixed with two moles of gas $B_{2}$ in a flask of volume 1lit. If at equilibrium 0.5 moles of $A_{2}$ is obtained. Then find out Kp for reaction $A_{2} (g) + B_{2} (g) \to 2 A B (g)$

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Solution

The correct option is D 36
$Initially At equilibrium {A_{2}}_{(g)} 2 0.5 + {B_{2}}_{(g)} 2 0.5 ⇌ 2 {A B}_{(g)} 0 2 \times 1.5 = 3$

Total moles at equilibrium $= 0.5 + 0.5 + 3 = 4 m o l$
Let, $P$ be the total pressure.
Therefore,
$K_{p} = \frac{{(\frac{3}{4} P)}^{2}}{(\frac{0.5}{4} P) (\frac{0.5}{4} P)}$

$\Rightarrow K_{p} = \frac{9}{0.25} = 36$

Hence $K_{p}$ for the given reaction is $36$ .

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Two moles of gas A2 are mixed with two moles of gas B2 in a flask of volume 1lit. If at equilibrium 0.5 moles of A2 is obtained. Then find out Kp for reaction A2(g)+B2(g)→2AB(g)

The correct option is D 36InitiallyAt equilibriumA2(g)20.5+B2(g)20.5⇌2AB(g)02×1.5=3Total moles at equilibrium =0.5+0.5+3=4 molLet, P be the total pressure.Therefore,Kp=(34P)2(0.54P)(0.54P)⇒Kp=90.25=36Hence Kp for the given reaction is 36.

Two moles of gas $A_{2}$ are mixed with two moles of gas $B_{2}$ in a flask of volume 1lit. If at equilibrium 0.5 moles of $A_{2}$ is obtained. Then find out Kp for reaction $A_{2} (g) + B_{2} (g) \to 2 A B (g)$