Derive the relation ${K_p} = {K_c}{(RT)^{\Delta ng}}$ for a general chemical equilibrium reaction.
Answer
583.2k+ views
Hint: We have to remember that in a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.
Complete answer:
We must remember that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.
Let the gaseous reaction is a state of equilibrium is:
$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$
Let ${p_A},{p_{B,}}{p_{C,}}{p_D}$ be the partial pressure of A, B, C and D respectively.
Therefore, we know that, the values of ${K_c}$ and ${K_p}$ are:
${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$ → (1)
${K_p} = \dfrac{{{p_c}^c{p_D}^d}}{{{p_A}^a{p_B}^b}}$ → (2)
From an ideal gas equation, we know that
PV= nRT
⇒P =$\dfrac{n}{V}$ RT = C R T
Here, C is referred to as the concentration.
Therefore,
${p_A}$= [A] R T
${p_B}$= [B] R T
${p_C}$= [C] R T
${p_D}$= [D] R T
Now, we will substitute the values in equation (2), so we get
${K_p} = \dfrac{{{{[C]}^c}{{(RT)}^c}{{[D]}^d}{{(RT)}^d}}}{{{{[A]}^a}{{(RT)}^a}{{[B]}^b}{{(RT)}^b}}}$
$ \Rightarrow {K_p} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}{(RT)^{[(c + d) - (a - b)]}}$
From eq (1).
$ \Rightarrow {K_p} = {K_c}{(RT)^{\Delta {n_g}}}$
Here,
$\Delta ng$ is the total no. of moles of gaseous product – the total no. of moles of gaseous a reactant. So, the relation between ${K_c}$ and ${K_p}$ is:
${K_p} = {K_c}{(RT)^{\Delta ng}}$ Hence, Proved.
Note:
In the end, we need to note that in equilibrium, the reactants and the products are present in two or more that two phases, it is called a heterogeneous equilibrium. Whereas, in an equilibrium when all the reactants and the products are present in the same phase it is called homogeneous equilibrium.
Complete answer:
We must remember that the equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.
Let the gaseous reaction is a state of equilibrium is:
$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$
Let ${p_A},{p_{B,}}{p_{C,}}{p_D}$ be the partial pressure of A, B, C and D respectively.
Therefore, we know that, the values of ${K_c}$ and ${K_p}$ are:
${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$ → (1)
${K_p} = \dfrac{{{p_c}^c{p_D}^d}}{{{p_A}^a{p_B}^b}}$ → (2)
From an ideal gas equation, we know that
PV= nRT
⇒P =$\dfrac{n}{V}$ RT = C R T
Here, C is referred to as the concentration.
Therefore,
${p_A}$= [A] R T
${p_B}$= [B] R T
${p_C}$= [C] R T
${p_D}$= [D] R T
Now, we will substitute the values in equation (2), so we get
${K_p} = \dfrac{{{{[C]}^c}{{(RT)}^c}{{[D]}^d}{{(RT)}^d}}}{{{{[A]}^a}{{(RT)}^a}{{[B]}^b}{{(RT)}^b}}}$
$ \Rightarrow {K_p} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}{(RT)^{[(c + d) - (a - b)]}}$
From eq (1).
$ \Rightarrow {K_p} = {K_c}{(RT)^{\Delta {n_g}}}$
Here,
$\Delta ng$ is the total no. of moles of gaseous product – the total no. of moles of gaseous a reactant. So, the relation between ${K_c}$ and ${K_p}$ is:
${K_p} = {K_c}{(RT)^{\Delta ng}}$ Hence, Proved.
Note:
In the end, we need to note that in equilibrium, the reactants and the products are present in two or more that two phases, it is called a heterogeneous equilibrium. Whereas, in an equilibrium when all the reactants and the products are present in the same phase it is called homogeneous equilibrium.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

