Question

For the following reaction, Kp = 1.9 ✕ 104 at 1722 K. H2(g) + Br2(g) equilibrium reaction arrow 2 HBr(g) What is the value of Kp for the following reactions at 1722 K? (a) HBr(g) equilibrium reaction arrow 1/2 H2(g) + 1/2 Br2(g)

Answer 1

Answer: The value of equilibrium constant for the given reaction is, $7.2\times 10^{-3}$

Explanation:

The given chemical equation is:

$H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)$; $K_p=1.9\times 10^4$

We need to calculate the equilibrium constant for the chemical equation, which is:

$HBr(g)\rightarrow \frac{1}{2}H_2(g)+\frac{1}{2}Br_2(g)$; $K_p'=?$

If the equation is revered then the equilibrium constant will be the reciprocal of the reaction.

If the equation is half then the equilibrium constant will be square-root of the reaction.

The value of equilibrium constant for the given reaction is:

$K_p'=(\frac{1}{K_p})^{1/2}$

$K_p'=(\frac{1}{1.9\times 10^4})^{1/2}$

$K_p'=7.2\times 10^{-3}$

Hence, the value of equilibrium constant for the given reaction is, $7.2\times 10^{-3}$