Question

a reaction mixture containing H2,N2 and NH3 has partial pressure 2 atm ,1atm and 3atm respectively at 725K . If the value of Kp for the reaction, N2 +3H2 gives 2Nh3 is 4.28 × 10^-2 atm^-2 at 725 k , in which direction the net reaction will go​

Answer 1

Answer: The net reaction is in backward direction.

Explanation:

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given equation:

$N_2+3H_2\rightleftharpoons 2NH_3$

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}$

We are given:

$p_{NH_3}=3atm\\p_{H_2}=2atm\\p_{N_2}=1atm$

Putting values in above equation, we get:

$Q_p=\frac{3^2}{(2)^3\times 1}=1.125$

We are given:

$K_p=4.28\times 10^{-2}$

There are 3 conditions:

• When $K_{p}>Q_p$; the reaction is product favored.
• When $K_{p}<Q_p$; the reaction is reactant favored.
• When $K_{p}=Q_p$; the reaction is in equilibrium.

As, $Q_p>K_p$, the reaction will be favoring reactant side.

Hence, the net reaction is in backward direction.