Question

At $1000K$ , the value of ${K_p}$ for the following reaction is $0.05atm$. The value of ${K_c}$ in terms of $R$ would be?
${A_{(g)}} + 2{B_{(g)}} \rightleftharpoons 3{C_{(g)}} + {D_{(g)}}$
A.$20000R$
B.$0.02R$
C.$5 \times {10^{ - 5}}R$
D.$5 \times {10^{ - 5}}{R^{ - 1}}$

Answer 1

We have to remember that the gaseous mixtures equilibrium constants are called ${K_p}$ & ${K_c}$ . When there is a reversible reaction (also called forward and reversed reaction), where the ratio of starting materials and products in the chemical reactions remains fixed, i.e. no net change in amount of products and reactants, $K$ is a equilibrium constant. At equilibrium, there is no driving force for either chemical reaction.

Complete step by step answer:
We must know that the present concentrations of reactants and products in chemical equilibrium is no further tendency to change with and the velocities for a reversible reaction is fixed, not changes.
${K_c}$ , molar concentration of a reaction, it is defined as the difference between the two constants in the reaction.${K_c}$ is used where the reaction is expressed in molarity. ${K_p}$ , is an equilibrium constant, used when the reaction is given as atmospheric pressure. These are given as gas equilibrium constants. The two equilibrium constants relation is given as,
${K_p} = {K_c}{(RT)^{\Delta n}}$
Where, $R$ is a universal gas constant, $T$ is temperature in kelvin and $\Delta n$ is the difference between the total number of moles in the reaction, i.e. the number of gaseous moles products and reactants.
Given reaction is,
${A_{(g)}} + 2{B_{(g)}} \rightleftharpoons 3{C_{(g)}} + {D_{(g)}}$
${K_p}$ for the following reaction is $0.05atm$ and T is $1000K$ .
$\Delta n = {\text{No}}{\text{. of moles of product(products - reactants)}}$
$\Rightarrow \Delta n = 4 - 3 = 1$
Substitute the given values in gas equilibrium constants relation we get,
$\Rightarrow 0.05 = {K_c}{(R \times 1000)^1}$
Rewrite the above equation, we get,
${K_c} = \dfrac{{0.05}}{{R \times 1000}}$
On simplification we get,
${K_c} = 5 \times {10^{ - 5}}{R^{ - 1}}$
From the above data, ${K_c}$ in terms of $R$ would be in the chemical reaction is $5 \times {10^{ - 5}}{R^{ - 1}}$ .
So, the correct answer is “Option D”.

Note:
We need to remember that the ${K_p}$ and ${K_c}$ , equilibrium constants, these constants may have units depending upon the nature of the reaction (even though thermodynamic equilibrium constant do not) and liquids & solids single component concentrations are not included.