Question

At ${{527}^{\circ }}C$the reaction has${{K}_{c}}=4$.
$N{{H}_{3}}\left( g \right)\dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{3}{2}{{H}_{2}}\left( g \right)$. So, what is the ${{K}_{p}}$of this reaction
${{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)2N{{H}_{3}}\left( g \right)$.

Answer 1

Hint: Since, the ${{K}_{c}}$of the reaction is given in the question. We can find the ${{K}_{p}}$using the formula which relates ${{K}_{P}},{{K}_{C}}$, change in number of moles of reactants and products, and temperature.
The formula is ${{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta {{n}_{g}}}}$.

Complete step-by-step answer:${{K}_{c}}$is known as the equilibrium constant when the concentration of the reactants and products are given in moles per liter.
For the reaction $\begin{align}
  & N{{H}_{3}}\left( g \right)\dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{3}{2}{{H}_{2}}\left( g \right) \\
 & {{K}_{c}}=4 \\
\end{align}$…… (equation 1)
Therefore, we can calculate ${{K}_{P}}$for this reaction using the formula
${{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta ng}}$… (equation 2)
Here, R= universal gas constant
T= temperature = ${{527}^{\circ }}C=527+273=800K$
$\Delta {{n}_{g}}=$(moles of products – moles of reactants)
$\Delta {{n}_{g}}=\left( \dfrac{3}{2}+\dfrac{1}{2} \right)-\left( 1 \right)=2-1=1$
Putting all these values in equation 1, we get
$\begin{align}
  & {{K}_{P}}=4{{\left( RT \right)}^{1}} \\
 & {{K}_{P}}=4RT \\
\end{align}$
Now, the another given reaction is ${{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)2N{{H}_{3}}$ … (equation 3)
We are asked to calculate ${{K}_{p}}$of the above reaction.
If we reverse the equation 1 and multiply with $2$, we get the above equation 3
If we have reversed the equation 1 then, the ${{K}_{p}}$will also be inversed. Also, we have multiplied the equation 1 with 2, so we will square the inversed ${{K}_{P}}$to obtain new one for the reaction represented by equation 3.
Let ${{K}_{P}}^{'}=$new ${{K}_{p}}$of equation 3
Therefore, $K_{P}^{'}=\dfrac{1}{{{K}_{P}}}$
$\begin{align}
  & K_{P}^{'}=\frac{1}{4\times R\times T} \\
 & K_{P}^{'}=\frac{1}{4R\times 800}=\frac{1}{3200R} \\
\end{align}$

Hence, this is the required solution for the question.

Note:Equilibrium constant only changes with the change in the temperature. You should always subtract the number of moles of reactants from the number of moles of products to calculate the value of . Students often do the reverse, that is they subtracts products moles from reactants mole. This makes the solution wrong..