Question

Calculate the following:
A.$\Delta {G^ \circ }$
B.The equilibrium constant for the formation of $N{O_2}$ from $NO$ and ${O_2}$ at 298 K.
$NO(g) + \dfrac{1}{2}{O_2}(g) \rightleftharpoons N{O_2}(g)$
Where ${\Delta _f}{G^ \circ }(N{O_2})$= 52.0 kJ/mol, ${\Delta _f}{G^ \circ }(NO)$= 87.0 kJ/mol and ${\Delta _f}{G^ \circ }({O_2})$= 0 kJ/mol.

Answer 1

There are certain things we have to keep in mind to solve his question which are as follows,
To calculate $\Delta {G^ \circ }$we can use the following formula,
$\Delta {G^ \circ }$(reaction) = $\Delta {G^ \circ }$ (products) - $\Delta {G^ \circ }$ (reactants)
and then, so as to calculate the equilibrium constant for the reaction ${K_c}$ we can use the following formula,
$\Delta {G^ \circ }$ = $- 2.303RT\log {K_c}$
Where, $\Delta {G^ \circ }$ is the free energy change of the reaction, R is the universal gas constant, T is the temperature at which reaction takes place and ${K_c}$ is the equilibrium constant for the reaction.

Complete step by step answer:
A.$NO(g) + \dfrac{1}{2}{O_2}(g) \rightleftharpoons N{O_2}(g)$
Applying the equation of change in Gibbs free energy at standard conditions and substituting the values, we have:
${\Delta _r}{G^ \circ } = \sum {{\Delta _f}{G^ \circ }_{products}} - \sum {{\Delta _f}{G^ \circ }_{reac\tan ts}}$
= ${\Delta _f}{G^ \circ }(N{O_2}) - ({\Delta _f}{G^ \circ }(NO) + \dfrac{1}{2}{\Delta _f}{G^ \circ }({O_2}))$
= 52.0 – (87.0 + $\dfrac{1}{2}$(0)) = 52 – 87 = -35 kJ/mol
As the value of Gibbs free energy is negative, this implies that the given reaction is spontaneous.

B.In order to determine the value of equilibrium constant, we need to use the relation between the change in Gibbs free energy at standard conditions and the equilibrium constant.
$\Delta {G^ \circ }$ = $- 2.303RT\log {K_c}$
Substituting the values, we have:
$- 35000{\text{ }} = {\text{ }} - 2.303 \times 8.314 \times 298 \times \log {K_c}$
Now, solving to get equilibrium constant, we have:
$log$ ${K_c}$= $\dfrac{{35000}}{{2.303 \times 8.314 \times 298}}$
$\log {K_c} = 6.314$
Taking antilog we get,
${K_c}$= $1.361 \times {10^6}$
Hence, the correct answers for the given question are $\Delta {G^ \circ }$= $- 35kJ/mol$ and ${K_c}$= $1.361 \times {10^6}$.

Note:
$\Delta {G^ \circ }$ can be defined as the change in free energy of a system as it goes from some initial state, such as all reactants, to some other, final state, such as all products. The value obtained for this thermodynamic variable tells us the maximum energy released which can be used or absorbed in going from the initial to the final state. For a reaction to proceed spontaneously the value of Gibbs free energy for that reaction should be negative.