Question

At a certain temperature, Kp = 4.5 x 103 for the reaction: 2H2S(g) = 2H2(g) + S2(g) If P[H2S] = 0.0050 atm, P[H2] = 0.50 atm, and P[S2] = 0.75 atm, then Qp > Kp and the reaction will shift toward products. O Qp > Kp and the reaction will shift toward reactants. O Qp <Kp and the reaction will shift toward products. O Qp = Kp and the reaction is at equilibrium. Qp <Kp and the reaction will shift toward reactants

Answer 1

1)The reaction could be written as-

The reaction quotient for this reaction can be written as-

2H S(g) = 2H2(g) + S2(g)

$Q_{P}=\frac{(p_{H_{2}})^2\times(p_{S_{2}})}{(p_{H_{2}S})^2}$......................................(i)

Given,

$P[H_{2}S]=0.0050atm$

$P[H_{2}]=0.50atm$

$P[S_{2}]=0.75atm$

Putting these values in equation-(i)

$Q_{P}=\frac{(0.50)^2(0.75)}{(0.0050)^2}=7.5\times10^3$

As the value of $Q_{P}$ is greater than the value of $K_{P}$ i.e.. indicates the reaction is not in equilibrium.Therefore in order to maintain the equilibrium the reaction would shifted towards right or reactants..

So the answer is-

$Q_{P}>K_{P}$ and the reaction will shift toward reactants.