QuestionJuly 29, 2025

For the reaction: P_(4)(s)+3H_(2)O(l)leftharpoons 2PH_(3)(g)+P_(2)O_(3)(g) At equilibrium, the reaction contains concentrations of: P_(4)=1.20M H_(2)O=4.50M PH_(3)=3.50M P_(2)O_(3)=1.50M Determine the value of the equilibrium constant K Answer

For the reaction: P_(4)(s)+3H_(2)O(l)leftharpoons 2PH_(3)(g)+P_(2)O_(3)(g) At equilibrium, the reaction contains concentrations of: P_(4)=1.20M H_(2)O=4.50M PH_(3)=3.50M P_(2)O_(3)=1.50M Determine the value of the equilibrium constant K Answer
For the reaction: P_(4)(s)+3H_(2)O(l)leftharpoons 2PH_(3)(g)+P_(2)O_(3)(g)
At equilibrium, the reaction contains concentrations of:
P_(4)=1.20M
H_(2)O=4.50M
PH_(3)=3.50M
P_(2)O_(3)=1.50M
Determine the value of the equilibrium constant K
Answer

Solution
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Answer

K_c = 0.168 Explanation 1. Write the expression for K_c The equilibrium constant K_c is given by the formula: **K_c = \frac products reactants **. For this reaction, K_c = \frac PH_3]^2[P_2O_3 P_4][H_2O]^3}. 2. Substitute concentrations into K_c expression Substitute the given concentrations into the K_c expression: K_c = \frac{(3.50)^2 \times 1.50}{1.20 \times (4.50)^3}. 3. Calculate K_c Calculate the value: K_c = \frac{12.25 \times 1.50}{1.20 \times 91.125} = \frac{18.375}{109.35}.

Explanation

1. Write the expression for $K_c$<br /> The equilibrium constant $K_c$ is given by the formula: **$K_c = \frac{[products]}{[reactants]}$**. For this reaction, $K_c = \frac{[PH_3]^2[P_2O_3]}{[P_4][H_2O]^3}$.<br />2. Substitute concentrations into $K_c$ expression<br /> Substitute the given concentrations into the $K_c$ expression: $K_c = \frac{(3.50)^2 \times 1.50}{1.20 \times (4.50)^3}$.<br />3. Calculate $K_c$<br /> Calculate the value: $K_c = \frac{12.25 \times 1.50}{1.20 \times 91.125} = \frac{18.375}{109.35}$.
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