Question

Be sure to answer all parts. The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine, 2NO(g) + Cl2(g) ⇌ 2NOCl(g) is 6.5 × 104 at 35°C. Calculate KP for this reaction, and determine whether the reaction will proceed to the right or to the left to achieve equilibrium when the starting pressures are PNO = 1.01 atm, PCl2 = 0.42 atm, and PNOCl = 1.76 atm.

Answer 1

Answer: The value of $K_p$ is $2.57\times 10^{3}$ and the reaction must proceed in the forward direction.

Explanation:

The given chemical equation follows:

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

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^{4}$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=35+273=308K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^{4}\times (0.0821\times 308)^{-1}\\\\K_p=2.57\times 10^{3}$

$K_p$ 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.

The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=1.76atm\\p_{NO}=1.01atm\\p_{Cl_2}=0.42atm$

Putting values in above equation, we get:

$Q_p=\frac{(1.76)^2}{(1.01)^2\times 0.42}=7.23$

We are given:

$K_p=2.57\times 10^3$

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

Hence, the value of $K_p$ is $2.57\times 10^{3}$ and the reaction must proceed in the forward direction.