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2P(s) + X(g) ↔ P2X(g) at 20°C, Keq= 4.0 Initially, 2.0 moles of P(s) and 1.0 mole of X(g) are put into a 1.0L flask and allowed to react until they reach equilibrium. The molarity of P2X (g) present at equilibrium is: a. 1.0 M b. 0.8 M c. 0.4 M d. 0.1 M

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Answer:

b. 0.8 M

Explanation:

Based on the reaction, Keq is defined as:

Keq = 4.0 = [P2X] / [X]

As P is a pure solid, its concentration is not taken into account

The initial concentration of X is 1.0mol/L = 1M, the equilibrium concentrations are:

[X] = 1M - X

[P2X] = X

Where X is the reaction coordinate

Replacing:

4.0 = X / 1-X

4.0 - 4.0X = X

4.0 = 5.0X

0.8M = X = [P2X]

Right answer is:

b. 0.8 M

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