133k views
4 votes
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.×10(Enter your answer in scientific notation.)reaction will proceed to the rightreaction is at equilibriumreaction will proceed to the left

User Wonsup Lee
by
8.0k points

1 Answer

3 votes

Answer: The reaction proceeds in the forward direction

Step-by-step explanation:

For the given chemical equation:


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

Relation of
K_p\text{ with }K_c is given by the formula:


K_p=K_c(RT)^(\Delta n_g)

where,


K_p = equilibrium constant in terms of partial pressure = ?


K_c = equilibrium constant in terms of concentration =
6.5* 10^4

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

T = temperature =
35^oC=[35+273]K=308K


\Delta n_g = 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* 10^4* (0.0821* 500)^(-1)\\\\K_p=1583.43


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=((p_(NOCl))^2)/(p_(Cl_2)* (p_(NO))^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=((1.76)^2)/(0.42* (1.01)^2)=7.23

We are given:


K_p=1583.43

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 reaction proceeds in the forward direction

User MattyP
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.