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Question: At \({{527}^{\circ }}C\)the reaction has\({{K}_{c}}=4\). \(N{{H}_{3}}\left( g \right)\dfrac{1}{2}...

At 527∘C{{527}^{\circ }}Cthe reaction hasKc=4{{K}_{c}}=4.
NH3(g)12N2(g)+32H2(g)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 Kp{{K}_{p}}of this reaction
N2(g)+3H2(g)2NH3(g){{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)2N{{H}_{3}}\left( g \right).

Explanation

Solution

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

Complete step-by-step answer: Kc{{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 NH3(g)12N2(g)+32H2(g) Kc=4 \begin{aligned} & 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{aligned}…… (equation 1)
Therefore, we can calculate KP{{K}_{P}}for this reaction using the formula
KP=KC(RT)Δng{{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta ng}}… (equation 2)
Here, R= universal gas constant
T= temperature = 527∘C=527+273=800K{{527}^{\circ }}C=527+273=800K
Δng=\Delta {{n}_{g}}=(moles of products – moles of reactants)
Δng=(32+12)−(1)=2−1=1\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
KP=4(RT)1 KP=4RT \begin{aligned} & {{K}_{P}}=4{{\left( RT \right)}^{1}} \\\ & {{K}_{P}}=4RT \\\ \end{aligned}
Now, the another given reaction is N2(g)+3H2(g)2NH3{{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)2N{{H}_{3}} … (equation 3)
We are asked to calculate Kp{{K}_{p}}of the above reaction.
If we reverse the equation 1 and multiply with 22, we get the above equation 3
If we have reversed the equation 1 then, the Kp{{K}_{p}}will also be inversed. Also, we have multiplied the equation 1 with 2, so we will square the inversed KP{{K}_{P}}to obtain new one for the reaction represented by equation 3.
Let KP′={{K}_{P}}^{'}=new Kp{{K}_{p}}of equation 3
Therefore, KP′=1KPK_{P}^{'}=\dfrac{1}{{{K}_{P}}}
KP′=14×R×T KP′=14R×800=13200R \begin{aligned} & K_{P}^{'}=\frac{1}{4\times R\times T} \\\ & K_{P}^{'}=\frac{1}{4R\times 800}=\frac{1}{3200R} \\\ \end{aligned}

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..