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Question: \({K_P}\) for the reaction \({N_2} + 3{H_2} \rightleftharpoons 2N{H_3}\) at \({400^o}C\) is \(1.64 \...

KP{K_P} for the reaction N2+3H2⇌2NH3{N_2} + 3{H_2} \rightleftharpoons 2N{H_3} at 400oC{400^o}C is 1.64×10−41.64 \times {10^{ - 4}} . Find KC{K_C} . Also find ΔGo\Delta {G^o} using KP{K_P} and KC{K_C}values and interpret the difference.

Explanation

Solution

he standard Gibbs free energy of formation of a compound is the change in Gibbs free energy that consists of the formation of one mole of that substance from its component elements, at their standard states which is the most stable form of the element at 25 ∘C25{\text{ }}^\circ C and100 kPa100{\text{ }}kPa . Its symbol is .

Complete answer:
KP{K_P} for the reaction is the equilibrium constant calculated from the partial pressure of the gases used in the reaction as reactants. As per the question, the reaction given is as follows:
N2+3H2⇌2NH3{N_2} + 3{H_2} \rightleftharpoons 2N{H_3}
The value of difference in the number of moles of product and reactants in the reaction will be:
Δng=np−nr\Delta {n_g} = {n_p} - {n_r}
Thus, the value of Δng\Delta {n_g} comes out to be:
Δng=2−4=−2\Delta {n_g} = 2 - 4 = - 2
KC{K_C} for the reaction is the equilibrium constant calculated from the concentration of the gases used in the reaction as reactants.
The relation between KP{K_P} and KC{K_C} is as follows:
KP=KC(RT)Δng{K_P} = {K_C}{(RT)^{\Delta {n_g}}}
Where, KP{K_P} for the reaction at 400oC{400^o}C= 1.64×10−41.64 \times {10^{ - 4}}
Δng=2−4=−2\Delta {n_g} = 2 - 4 = - 2
KC=?{K_C} = ?
R=0.082LatmK−1mol−1R = 0.082Latm{K^{ - 1}}mo{l^{ - 1}}
T=400oC=(400+273)K=673KT = {400^o}C = (400 + 273)K = 673K
Substituting the values and solving, we have:
1.64×10−4=KC(0.082×673)−21.64 \times {10^{ - 4}} = {K_C}{(0.082 \times 673)^{ - 2}}
⇒KC=1.64×10−4×(0.082×673)2≈0.50\Rightarrow {K_C} = 1.64 \times {10^{ - 4}} \times {(0.082 \times 673)^2} \approx 0.50
Thus the value of KC{K_C} is found to be 0.500.50.
The relation between equilibrium constant and change in standard Gibbs free energy is:
ΔGo=−RTln⁡K\Delta {G^o} = - RT\ln K
Now, substituting the values of KP{K_P} and KC{K_C}separately, we get:
ΔGo=−RTln⁡KP\Delta {G^o} = - RT\ln {K_P}….(i)
ΔGo=−RTln⁡KC\Delta {G^o} = - RT\ln {K_C}….(ii)
Substituting the values in equation (i) and (ii) and on solving, we have:
ΔGo=−8.314×673×ln⁡(1.64×10−4)=−8.314×673×(−3.506)=19617.19J/mol\Delta {G^o} = - 8.314 \times 673 \times \ln (1.64 \times {10^{ - 4}}) = - 8.314 \times 673 \times ( - 3.506) = 19617.19J/mol
The above value is for the change in standard Gibbs free energy, when the equilibrium is taken for the partial pressure of the reactants.
ΔGo=−8.314×673×ln⁡(0.50)=−8.314×673×(−0.693)=3877.55J/mol\Delta {G^o} = - 8.314 \times 673 \times \ln (0.50) = - 8.314 \times 673 \times ( - 0.693) = 3877.55J/mol
The above value is for the change in standard Gibbs free energy, when the equilibrium is taken for the concentration of the reactants.
The standard free energy change calculated form KP{K_P} is higher than the standard free energy change calculated from KC{K_C} ​as the numerical value of KC{K_C}​ is higher than the numerical value of KP{K_P}.

Note:
The value of the universal gas constant should always be decided on the basis of the values of the other variables given in the question and not any random values. In the above reaction of formation of ammonia from nitrogen and hydrogen, the value of KP{K_P} is less than the value of KC{K_C}.