Question

For the reaction: 2 A (g) + B (s)= 2 C(s) + D (g)
At 298 K in a 10.0 L vessel, the equilibrium values are as follows: 0.721 atm of A, 4.18 mol of B, 6.25 mol of C, and 2.71 atm of D. What is the value of the equilibrium constant?

Answer 1

Answer:

The value of the equilibrium constant = 5.213

Explanation:

Here $K_p$ (equilibrium constant) is referred to as the  partial pressure of product divided by the  partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation .

As such only gas appear in $K_p$  expression as solids takes a value of 1;

SO ; in the  given equation from the question:

2 A (g) + B (s) ----> 2 C(s) + D (g)

$K_p = \dfrac{[D]}{[A]^2}$

$K_p = \dfrac{2.71}{0.721^2}$

$K_p = 5.213$

The value of the equilibrium constant = 5.213

Answer 2

The value of the equilibrium constant is 5.213.

Given that,

• The reaction: 2 A (g) + B (s)= 2 C(s) + D (g)
• At 298 K in a 10.0 L vessel, the equilibrium values are as follows: 0.721 atm of A, 4.18 mol of B, 6.25 mol of C, and 2.71 atm of D.

Based on the above information, the calculation is as follows:

$= 2.71 \div 0.721^2$

= 5.213

Therefore we can conclude that the value of the equilibrium constant is 5.213.

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