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Question: If \({K_P}\) for the given reaction is \(1.44 \times {10^{ - 5}}units\) , then the value of \({K_C}\...

If KP{K_P} for the given reaction is 1.44×10−5units1.44 \times {10^{ - 5}}units , then the value of KC{K_C} will be :
N2(g)+3H2(g)⟷2NH3(g){N_2}\left( g \right) + 3{H_2}\left( g \right)\underset{{}}{\overset{{}}{\longleftrightarrow}}2N{H_3}\left( g \right)
a.) 1.44×10−5(0.082×500)mol−2L2\dfrac{{1.44 \times {{10}^{ - 5}}}}{{\left( {0.082 \times 500} \right)}}mo{l^{ - 2}}{L^2}
b.) 1.44×10−5(8.314×773)mol2L−2\dfrac{{1.44 \times {{10}^{ - 5}}}}{{\left( {8.314 \times 773} \right)}}mo{l^2}{L^{ - 2}}
c.) 1.44×10−5(0.082×773)mol2L−2\dfrac{{1.44 \times {{10}^{ - 5}}}}{{\left( {0.082 \times 773} \right)}}mo{l^2}{L^{ - 2}}
d.) 1.44×10−5(0.082×773)mol−2L2\dfrac{{1.44 \times {{10}^{ - 5}}}}{{\left( {0.082 \times 773} \right)}}mo{l^{ - 2}}{L^2}

Explanation

Solution

The reaction given to us occurs at T=500∘C=500+273=773KT = 500^\circ C = 500 + 273 = 773K . We know that the relationship between Kp{K_p} and Kc{K_c} is KC=Kp(RT)Δn{K_C} = \dfrac{{{K_p}}}{{{{\left( {RT} \right)}^{\Delta n}}}} . Use this equation to calculate the value of Kc{K_c} .

Complete step by step answer:
The reaction given in the problem is the reaction involved in Haber’s process.
The Haber process also called the Haber-Bosch process, is an artificial process of fixation of nitrogen and is today the main industrial process for ammonia processing. It is named after its inventors, Fritz Haber and Carl Bosch, both German chemists, who invented it in the first decade of the 20th century. The method converts atmospheric nitrogen into ammonia through a hydrogen reaction using a metal catalyst at high temperatures and pressures.
The reaction given in the problem is
N2(g)+3H2(g)⟷2NH3(g){N_2}\left( g \right) + 3{H_2}\left( g \right)\underset{{}}{\overset{{}}{\longleftrightarrow}}2N{H_3}\left( g \right)
This reaction occurs at 500∘C500^\circ C
So, T=500∘C=500+273=773KT = 500^\circ C = 500 + 273 = 773K

The equilibrium constant - The chemical reaction equilibrium constant (usually denoted by the symbol K) provides insight into the product-reactant relationship when a chemical reaction reaches equilibrium.
For a simple reaction
aA+bB⇆cC+dDaA + bB\overset {} \leftrightarrows cC + dD
K=[C]c[D]d[A]a[B]bK = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}
Now, Kc={K_c} = Equilibrium constant when measured in moles per liter
Kp={K_p} = Equilibrium constant when calculated from the partial pressures
The relationship between Kp{K_p} and Kc{K_c} is as follows
KC=Kp(RT)Δn{K_C} = \dfrac{{{K_p}}}{{{{\left( {RT} \right)}^{\Delta n}}}}
Here, R=0.082atmL/molK=R = 0.082atmL/molK = Universal Gas Constant
Δn=\Delta n = The difference in the no. of moles between the products and the reactants
So the equation for this problem becomes
KC=1.44×10−5(0.0.821×773)−2mol−2L2{K_C} = \dfrac{{1.44 \times {{10}^{ - 5}}}}{{{{\left( {0.0.821 \times 773} \right)}^{ - 2}}}}mo{l^{ - 2}}{L^2}
So, the correct answer is “Option D”.

Note: Ammonia was difficult to manufacture on an industrial scale before the invention of the Haber process, with early techniques such as the Birkeland-Eyde process and the Frank-Caro process both being extremely inefficient. The Haber method is primarily used today to produce ammonia-based fertilizer.