Question

The reaction between ${{N}_{2}}$ and ${{H}_{2}}$ to form ammonia has ${{K}_{c}} = 6\times {{10}^{-2}}$ at the temperature ${{500}^{\circ }}C$. The numerical value of for this reaction is:
A. $1.5\times {{10}^{-5}}$
B. $1.5\times {{10}^{5}}$
C. $1.5\times {{10}^{-6}}$
D. $1.5\times {{10}^{6}}$

Answer 1

Hint: We have to find the value of ${{K}_{p}}$ in this question. For that we need to first convert the unit of temperature into kelvin and find out the value of $\Delta n$. We will find ${{K}_{p}}$ by the formula: ${{K}_{p}}={{K}_{c}}\times {{(RT)}^{\Delta n}}$

Complete Solution :
 - We can write the reaction between ${{N}_{2}}$ and ${{H}_{2}}$ to form ammonia as:
${{N}_{2}}+3{{H}_{2}}\rightleftarrows 2N{{H}_{3}}$

- We are being provided with the value of ${{K}_{c}} = 6\times {{10}^{-2}}$
- We will convert the temperature that was given in Celsius into kelvin. We can write it as:
T = ${{500}^{\circ }}C$ = 500 + 273K = 773K

- Now, proceeding further we will find the value of $\Delta n$ = total number of moles of gaseous products - total number of moles of gaseous reactants
$\Delta n = 2 - (1 + 3) = -2$
- Now, we will find the value of ${{K}_{p}}$ by the formula:
${{K}_{p}}={{K}_{c}}\times {{(RT)}^{\Delta n}}$

- Now, by putting all the values in the given formula we get:
${{K}_{p}} = 6\times {{10}^{-2}}\times {{(0.08206L.atm/mol.K\times 773K)}^{-2}}$
${{K}_{p}}$ = $1.5\times {{10}^{-5}}$
- Hence, we can conclude that the correct option is (a), that is the numerical value of this reaction is $1.5\times {{10}^{-5}}$.
So, the correct answer is “Option A”.

Note: - As we know that and are both ${{K}_{c}}$ and ${{K}_{p}}$ equilibrium constant of an ideal gas. But the only difference in both is that ${{K}_{p}}$ is used when equilibrium concentrations are expressed in pressure.
- Whereas, ${{K}_{c}}$ is used when equilibrium concentrations are expressed in molarity.
- It is found that when there is change in the number of moles of gas is zero, or we can say ${{K}_{p}}$ =${{K}_{c}}$