For a reaction A ( g ) ⇌ 2 B ( g ) , the initial pressure of A at 300 ∘ C is 1 atm and its K P is 2 atm. Then calculate P B at the equilibrium state.

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For a reaction $A (g) ⇌ 2 B (g)$ , the initial pressure of A at 300 $^{\circ} C$ is 1 atm and its $K_{P}$ is 2 atm. Then calculate $P_{B}$ at the equilibrium state.

Options:

Answer:

Solution:

Let x atm of A react: $P_{A} = 1 - x, P_{B} = 2 x$ .

$K_{P} = (2 x)^{2} / (1 - x) = 2$ , so $4 x^{2} = 2 - 2 x$ and $2 x^{2} + x - 1 = 0$ , i.e. (2x - 1)(x + $1) = 0$ , so $x = 0.5$ .

$P_{B} = 2 x = 1.0$ atm.

Trap: 0.5 atm is × (and $P_{A}$ ), not $P_{B}$ .

Answer: (B).

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp

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