Question

For the reaction, ${X_2}{O_{4(i)}} \to 2X{O_{2(g)}},{\text{ }}\Delta U = 2.1kcal,{\text{ }}\Delta S = 20cal{K^{ - 1}}$ at 300K. Hence $\Delta G$ is:
(A) 2.7 kcal
(B) -2.7 kcal
(C) 9.3 kcal
(D) -9.3 kcal

Answer 1

Below are the two equations of thermodynamic parameters which will lead to the answer.
$\Delta H = \Delta U + \Delta {n_g}RT{\text{ }}$
$\Delta G = \Delta H - T\Delta S{\text{ }}$

Complete answer:
We are given the internal energy change and the change in entropy during the reaction. We need to find the free energy change during the reaction. We will first find the change in enthalpy and then we will use it to find the change in free energy.
- We know that we can write enthalpy change in terms of internal energy as
$\Delta H = \Delta U + P\Delta V$
Now, according the ideal gas equation , we know that PV = nRT, so we can write the above equation as
$\Delta H = \Delta U + \Delta {n_g}RT{\text{ }}.....{\text{(1)}}$
Where $\Delta H$ is the change in enthalpy,
$\Delta U$ is the change in internal energy of the system (2.1kcal),
$\Delta {n_g}$ is the difference in molecules of gases between products and reactants,
R is the universal gas constant ($2 \times {10^{ - 3}}kcal \cdot mo{l^{ - 1}}{K^{ - 1}}$) and
T is temperature given as 300K.
Here, $\Delta {n_g}$ = number of moles of gases in products – number of moles of gases in reactants
$\Delta {n_g}$ = 2 – 0 = 2 moles
- So, putting all the available values into equation (1), we get
$\Delta H = 2.1 + (2)(2 \times {10^{ - 3}})(300)$
Thus, we obtained that
$\Delta H = 2.1 + 1.2 = 3.3kcal$
Now, we know that
$\Delta G = \Delta H - T\Delta S{\text{ }}..{\text{(2)}}$
We obtained that $\Delta H = 3.3kcal$ ,
T is 300 K and $\Delta S$ is given $20cal{K^{ - 1}}$ = $0.02kcal{K^{ - 1}}$
So, putting all these values into equation (2), we get
$\Delta G = 3.3 - (300)(0.02)$
So,
$\Delta G = 3.3 - 6 = - 2.7kcal$
Thus, we obtained that the free energy change of this reaction will be -2.7kcal.

So, the correct answer is (B).

Note:
Do not forget that there is a minus sign in equation (2). Remember that the value of the universal gas constant in $kcal \cdot mo{l^{ - 1}}{K^{ - 1}}$ unit is $2 \times {10^{ - 3}}$ . Here, it will be easy for us if we just use the universal gas constant in $kcal \cdot mo{l^{ - 1}}{K^{ - 1}}$ unit because other quantities are given in kcal units.