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Question: What is \[{\Delta _r}G\] (KJ / mole) for synthesis of ammonia at 298 K at following sets of partial ...

What is ΔrG{\Delta _r}G (KJ / mole) for synthesis of ammonia at 298 K at following sets of partial pressure:
N2(g)+3H2(g)⇌2NH3(g);ΔrG0=−33KJmol{N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g);{\Delta _r}{G^0} = - 33\dfrac{{KJ}}{{mol}}
[Take R = 8.3 J / K mole, log2=0.3log2 = 0.3; log3=0.48log3 = 0.48 ]

Gas:N2{N_2}H2{H_2}NH3N{H_3}
Pressure (atm):130.02

A.+6.5
B.-6.5
C.+60.5
D.-60.5

Explanation

Solution

For the given reaction, N2(g)+3H2(g)⇌2NH3(g);{N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g); the partial pressures of the reactant and product are given, also given are the ΔrG0{\Delta _r}{G^0}, Temperature (T) and R. So, substituting the value in the Gibbs free energy equation, ΔrG=ΔrG0+RTln QR{\Delta _r}G = {\Delta _r}{G^0} + RTln\,{Q_R} we can calculate the value of ΔrG{\Delta _r}G.

Complete step by step solution:
N2(g)+3H2(g)⇌2NH3(g){N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g) {Given reaction}
Given in the question are,
ΔrG0=−33KJmol{\Delta _r}{G^0} = - 33\dfrac{{KJ}}{{mol}}
R=8.3JK moleR = 8.3\dfrac{J}{{K\,mole}}
Temperature, T=298KT = 298K
Partial pressure of NH3N{H_3} gas pNH3=0.02{p_{N{H_3}}} = 0.02
Partial pressure of H2{H_2} gas pH2=3{p_{{H_2}}} = 3
Partial pressure of N2{N_2} gas pN2=1{p_{{N_2}}} = 1
The equation for Gibbs Free energy is,
ΔrG=ΔrG0+RTln⁡ QR{\Delta _r}G = {\Delta _r}{G^0} + RT\ln \,{Q_R} ………… Equation (1)
For a gas phase reaction of the type
aA(g)+bB(g)⇌cC(g)+dD(g)aA\left( g \right) + bB\left( g \right) \rightleftharpoons cC\left( g \right) + dD\left( g \right),
QR=pcC×pdDpaA×pbB{Q_R} = \dfrac{{{p^c}C \times {p^d}D}}{{{p^a}A \times {p^b}B}}
⇒ΔrG=ΔrG0+RTlnpcC×pdDpaA×pbB\Rightarrow {\Delta _r}G = {\Delta _r}{G^0} + RTln\dfrac{{{p^c}C \times {p^d}D}}{{{p^a}A \times {p^b}B}}
This shows that if you increase the partial pressure of a product gas, ΔG\Delta G becomes more positive.
If you increase the partial pressure of a reactant gas, ΔG\Delta G becomes more negative.

Now, QR=p2NH3p3H2×pN2{Q_R} = \dfrac{{{p^2}N{H_3}}}{{{p^3}{H_2} \times p{N_2}}}
=(0.02)232×1=0.00049=4.445= \dfrac{{{{(0.02)}^2}}}{{{3^2} \times 1}} = \dfrac{{0.0004}}{9} = 4.445
Substituting the values of ΔrG0{\Delta _r}{G^0}, QR{Q_R}, R and T in equation (1)
ΔrG=ΔrG0+RTln QR{\Delta _r}G = {\Delta _r}{G^0} + RTln\,{Q_R}
⇒\Rightarrow ΔrG=−33+8.3×298ln4.445{\Delta _r}G = - 33 + 8.3 \times 298ln4.445
⇒\Rightarrow ΔrG=−60.5KJmole{\Delta _r}G = - 60.5\dfrac{{KJ}}{{mole}}
So, the value of ΔrG=−60.5KJmole{\Delta _r}G = - 60.5\dfrac{{KJ}}{{mole}} for the synthesis of ammonia at 298 K.

Therefore, the correct answer is option (D).

Note: The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, which makes the reaction spontaneous only at low temperatures. Thus, higher T which speeds up the reaction also reduces the extent.