Question

For reaction, $ 2{\text{NOC}}{{\text{l}}_{({\text{g}})}} \leftrightharpoons 2{\text{N}}{{\text{O}}_{({\text{g}})}} + {\text{C}}{{\text{l}}_{2(\;{\text{g}})}} $ , $ {K_C}{\text{ at }}{427^\circ }{\text{C}} $ is $ 3 \times {10^{ - 6}}{\text{Lmo}}{{\text{l}}^{ - 1}} $ . The value of $ {{\text{K}}_{\text{p}}} $ is:
A) $ 7.50 \times {10^{ - 5}} $
B) $ 2.50 \times {10^{ - 5}} $
C) $ 2.50 \times {10^{ - 4}} $
D) $ 1.75 \times {10^{ - 4}} $

Answer 1

Hint :We use gas equilibrium constant to solve the problem. Pressure is exactly proportional to concentration, according to the ideal gas equation, provided volume and temperature are constant. We can formulate our equilibrium expression for a gas-phase reaction in terms of the partial pressures of each gas since pressure is directly proportional to concentration. This distinct equilibrium constant is referred to as $ {K_p} $ .

Complete Step By Step Answer:
However, for gas-specific reactions, the equilibrium constant may also be expressed in terms of the partial pressures of the gases involved. Consider a simple gas-phase process.
 $ aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) $
Our partial pressure equilibrium constant, denoted $ {K_p} $ , is as follows.
 $ {K_P} = \dfrac{{P_C^cP_D^d}}{{P_A^aP_B^b}} $
It's worth noting how close this expression is to $ {K_c} $ , the equilibrium expression stated in terms of concentrations. When composing $ {K_p} $ expressions, avoid using brackets ([ ]) to avoid misunderstanding.
By looking at the ideal gas law, we can see why we are permitted to formulate a K equation in terms of partial pressures for gases. Remember that the ideal gas law is defined as follows:
 $ PV = nRT $
We have rewritten this phrase in terms of P.
 $ P = \dfrac{n}{V}RT $
It's worth noting that in order for K to be constant, temperature must also be constant. As a result, the term RT in the preceding formulation is a constant. $ \dfrac{n}{V} $ (moles per unit volume) is just a concentration measurement. Because pressure is precisely proportional to concentration, we may safely utilise $ {K_p} $ .
Finally, there is a crucial equation that connects $ {K_p} $ and $ {K_c} $ . The following is the information.
 $ {K_p} = {K_c}{(RT)^{\Delta n}} $
Now using the formula
 $ 2{\text{NOC}}{{\text{l}}_{({\text{g}})}} \leftrightharpoons 2{\text{N}}{{\text{O}}_{({\text{g}})}} + {\text{C}}{{\text{l}}_{2(\;{\text{g}})}} $
 $ {\mathbf{\Delta n}} = {{\mathbf{n}}_{\mathbf{P}}} - {{\mathbf{n}}_{\mathbf{R}}} = {\mathbf{3}} - {\mathbf{2}} = {\mathbf{1}} $
Given:-
 $ {{\mathbf{K}}_{\text{C}}} = 3 \times {10^{ - 6}}{\text{Lmo}}{{\text{l}}^{ - 1}} $
 $ {\text{T}} = {427^\circ }{\text{C}} = (427 + 273){\text{K}} = 700\;{\text{K}} $
So know that,
 $ {{\mathbf{K}}_{\mathbf{P}}} = {{\mathbf{K}}_{\mathbf{C}}}{({\mathbf{RT}})^{\Delta {\mathbf{n}}}} $
 $ {{\mathbf{K}}_{\text{P}}} = 3 \times {10^{ - 6}}{({\mathbf{0}}.{\mathbf{0821}} \times {\mathbf{700}})^1} $
 $ {{\mathbf{K}}_{\text{P}}} = 1.72 \times {10^{ - 4}} \approx 1.75 \times {10^{ - 4}} $
 $ \Rightarrow {{\mathbf{K}}_{\text{P}}} = 1.75 \times {10^{ - 4}} $
Hence option D is correct.

Note :
The equilibrium constants of an ideal gaseous mixture are $ {K_p} $ and $ {K_c} $ . When equilibrium concentrations are represented in atmospheric pressure, $ {K_p} $ is the equilibrium constant to use, and when equilibrium concentrations are represented in molarity, $ {K_c} $ is the equilibrium constant to use.