Question

Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the equilibrium constant Kp for this reaction is 0.26. Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium.(a) PNO = 0.16 atm, PCl2 = 0.30 atm, and PNOCl = 0.11 atm.(b) PNO = 0.12 atm, PCl2 = 0.10 atm, and PNOCl = 0.048 atm.(c) PNO = 0.15 atm, PCl2 = 0.15 atm, and PNOCl = 5.20 10-3 atm.

Answer 1

Answer:

For a: The mixture will need to produce more reactants to reach equilibrium.

For b: The mixture will need to produce more reactants to reach equilibrium.

For c: The mixture will need to produce more products to reach equilibrium.

Explanation:

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

The expression of $K_{p}$ for above equation follows:

$K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$   .....(1)

We are given:

Value of $K_p$ = 0.26

There are 3 conditions:

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

For the given options:

• For a:

We are given:

$p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57$

As, $K_{p}$; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

• For b:

We are given:

$p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6$

As, $K_{p}$; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

• For c:

We are given:

$p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm$

Putting values in expression 1, we get:

$Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008$

As, $K_{p}>Q_p$; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.