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* 10^(3)` and the reaction must proceed in the forward direction.

Step-by-step 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* 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* 10^(4)* (0.0821* 308)^(-1)\\\\K_p=2.57* 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=((p_(NOCl))^2)/((p_(NO))^2* 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=((1.76)^2)/((1.01)^2* 0.42)=7.23`

We are given:

`K_p=2.57* 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* 10^(3)` and the reaction must proceed in the forward direction.