First slide

Chemical Equilibrium

Question

An arbitrary compound $P_{2} Q$ decomposes according to reaction:
$2 P_{2} Q (g) ⇌_{}^{} 2 P_{2} (g) + {Q_{2}}_{(g)}$
If one starts the decomposition reaction with 4 moles of $P_{2} Q$ and value of equilibrium constant $K_{p}$ is numerically equal to total pressure at equilibrium. Then which option is/are correct at equilibrium,.

Moderate

A

Moles of ${n_{P}}_{_{2} Q} = n_{Q_{2}}$
B

Moles of $n_{P_{2}} = (8 / 3)$
C

Degree of dissociation is $α = (2 / 3)$
D

Total number of moles of products $(P_{2} & Q_{2})$ at equilibrium is 4

Solution

$\begin{array}{l} 2 P_{2} Q ⇌_{}^{} 2 P_{2} + Q_{2} \\ t = 0 4 0 0 \\ t = e q 4 (1 - α) 4 α 2 α \end{array}$
$K_{p} = \frac{{(\frac{4 α}{4 + 2 α} \times P)}^{2} (\frac{2 α}{4 + 2 α} \times P)}{{(\frac{4 (1 - α)}{4 + 2 α} \times P)}^{2}} = \frac{2 α^{3}}{{(1 - α)}^{2} (4 + 2 α)} \times P$
But, $K_{p} = P (given) ∴ 2 α^{3} = (1 - α)^{2} (4 + 2 α)$
$∴ - 6 α + 4 = 0 ∴ α = (2 / 3)$
Total moles at eq $= 4 + 2 α = (16 / 3)$
$n_{P_{2} Q} = n_{Q_{2}} = (4 / 3), n_{P_{2}} = (8 / 3)$
\ Total number of moles of products $= \frac{8}{3} + \frac{4}{3} = \frac{12}{3} = 4.$

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