Question

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

Gas:	${N_2}$	${H_2}$	$N{H_3}$
Pressure (atm):	1	3	0.02

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

Answer 1

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

Complete step by step solution:
 ${N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)$ {Given reaction}
Given in the question are,
 \[{\Delta _r}{G^0} = - 33\dfrac{{KJ}}{{mol}}\]
 \[R = 8.3\dfrac{J}{{K\,mole}}\]
Temperature, \[T = 298K\]
Partial pressure of $N{H_3}$ gas ${p_{N{H_3}}} = 0.02$
Partial pressure of ${H_2}$ gas ${p_{{H_2}}} = 3$
Partial pressure of ${N_2}$ gas ${p_{{N_2}}} = 1$
The equation for Gibbs Free energy is,
 \[{\Delta _r}G = {\Delta _r}{G^0} + RT\ln \,{Q_R}\] ………… Equation (1)
For a gas phase reaction of the type
 \[aA\left( g \right) + bB\left( g \right) \rightleftharpoons cC\left( g \right) + dD\left( g \right)\],
 \[{Q_R} = \dfrac{{{p^c}C \times {p^d}D}}{{{p^a}A \times {p^b}B}}\]
 \[ \Rightarrow {\Delta _r}G = {\Delta _r}{G^0} + RTln\dfrac{{{p^c}C \times {p^d}D}}{{{p^a}A \times {p^b}B}}\]
This shows that if you increase the partial pressure of a product gas, \[\Delta G\] becomes more positive.
If you increase the partial pressure of a reactant gas, \[\Delta G\] becomes more negative.

Now, \[{Q_R} = \dfrac{{{p^2}N{H_3}}}{{{p^3}{H_2} \times p{N_2}}}\]
 $ = \dfrac{{{{(0.02)}^2}}}{{{3^2} \times 1}} = \dfrac{{0.0004}}{9} = 4.445$
Substituting the values of \[{\Delta _r}{G^0}\], \[{Q_R}\], R and T in equation (1)
 \[{\Delta _r}G = {\Delta _r}{G^0} + RTln\,{Q_R}\]
$\Rightarrow$ \[{\Delta _r}G = - 33 + 8.3 \times 298ln4.445\]
$\Rightarrow$ \[{\Delta _r}G = - 60.5\dfrac{{KJ}}{{mole}}\]
So, the value of \[{\Delta _r}G = - 60.5\dfrac{{KJ}}{{mole}}\] 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.