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* 10^(-3)`

Step-by-step explanation:

The given chemical equation is:

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

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

`HBr(g)\rightarrow (1)/(2)H_2(g)+(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'=((1)/(K_p))^(1/2)`

`K_p'=((1)/(1.9* 10^4))^(1/2)`

`K_p'=7.2* 10^(-3)`

Hence, the value of equilibrium constant for the given reaction is,
`7.2* 10^(-3)`