Answer to Question #145836 in General Chemistry for Angelo Torregoza

Question

Consider the reaction at 500K: H2(g) + Br2(g) ⇋ 2HBr(g). The molar concentration of hydrogen, bromine, and hydrogen bromide at equilibrium was found to be 0.25M, 0.25M, and 5.06M respectively. Compute for the value of the equilibrium constant Kp.

Answer 1

Firstly,we have to write the equilibrium constant equation for the reaction.

So we have;

K=[HBr]^2/[H2][BR2]

To find Kp, we need to find Kc since we were given concentration not pressure

So Kc=[5.06]^2/[0.25][0.25]

Kc=409.66

To get Kp we will use the formula;

Kp=Kc(RT)^dn

Where R is ideal gas constant which is 0.8206

T=Temperature in kelvin

dn=sum of the coefficient of the product minus that of the reactants which is

2-(1+1)=0

Inputing the values, we have;

Kp=409.66×(0.8206×500)^0

According to the laws of indices,anything raised to the power of zero is 1

So we have Kp=409.66×1

Kp=409.66