supatchaaue
Answered

Calculate the standard free energy change of these reactions.

i. [tex]\[ \text{Reaction: } SiO_2(s) + 4HF(g) \rightarrow SiF_4(g) + 2H_2O(l) \][/tex]
Standard free energy of formation values:
[tex]\[ \Delta G^{\ominus}[SiO_2(s)] = -856.7 \, \text{kJ mol}^{-1} \][/tex]
[tex]\[ \Delta G^{\ominus}[HF(g)] = -273.2 \, \text{kJ mol}^{-1} \][/tex]
[tex]\[ \Delta G^{\ominus}[SiF_4(g)] = -1572.7 \, \text{kJ mol}^{-1} \][/tex]
[tex]\[ \Delta G^{\ominus}[H_2O(l)] = -69.9 \, \text{kJ mol}^{-1} \][/tex]

ii. [tex]\[ \text{Reaction: } SrCO_3(s) \rightarrow SrO(s) + CO_2(g) \][/tex]
Standard free energy of formation values:
[tex]\[ \Delta G^{\ominus}[SrCO_3(s)] = -1140.4 \, \text{kJ mol}^{-1} \][/tex]
[tex]\[ \Delta G^{\ominus}[SrO(s)] = -561.9 \, \text{kJ mol}^{-1} \][/tex]
[tex]\[ \Delta G^{\ominus}[CO_2(g)] = -394.4 \, \text{kJ mol}^{-1} \][/tex]



Answer :

GinnyAnswer
To calculate the standard free energy change ([tex]\(\Delta G^\ominus\)[/tex]) of the given reactions, we'll use the standard free energy of formation ([tex]\(\Delta G_f^\ominus\)[/tex]) values provided.

The standard free energy change for a reaction can be calculated using the following equation:
[tex]\[ \Delta G^\ominus = \sum \Delta G_f^\ominus \text{(products)} - \sum \Delta G_f^\ominus \text{(reactants)} \][/tex]

Let's apply this formula to each reaction.

### Reaction i
Reaction:
[tex]\[ SiO_2 (s) + 4 HF (g) \rightarrow SiF_4 (g) + 2 H_2O (l) \][/tex]

Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SiO_2 (s)] = -856.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [HF (g)] = -273.2 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SiF_4 (g)] = -1572.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [H_2O (l)] = -69.9 \text{ kJ/mol}\)[/tex]

Calculation:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ \Delta G_f^\ominus(\text{SiF}_4) + (2) \Delta G_f^\ominus(\text{H}_2\text{O}) \right] - \left[ \Delta G_f^\ominus(\text{SiO}_2) + (4) \Delta G_f^\ominus(\text{HF}) \right] \][/tex]

Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 + 2(-69.9) \right] - \left[ -856.7 + 4(-273.2) \right] \][/tex]

[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 - 139.8 \right] - \left[ -856.7 - 1092.8 \right] \][/tex]

[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 - (-1949.5) \][/tex]

[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 + 1949.5 = 237 \text{ kJ/mol} \][/tex]

### Reaction ii
Reaction:
[tex]\[ SrCO_3 (s) \rightarrow SrO (s) + CO_2 (g) \][/tex]

Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SrCO_3 (s)] = -1140.4 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SrO (s)] = -561.9 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [CO_2 (g)] = -394.4 \text{ kJ/mol}\)[/tex]

Calculation:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ \Delta G_f^\ominus(\text{SrO}) + \Delta G_f^\ominus(\text{CO}_2) \right] - \left[ \Delta G_f^\ominus(\text{SrCO}_3) \right] \][/tex]

Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ -561.9 + (-394.4) \right] - \left[ -1140.4 \right] \][/tex]

[tex]\[ \Delta G^\ominus(\text{ii}) = (-561.9 - 394.4) - (-1140.4) \][/tex]

[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 - (-1140.4) \][/tex]

[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 + 1140.4 = 184.1 \text{ kJ/mol} \][/tex]

Hence, the standard free energy changes for the reactions are:
- For reaction i: [tex]\(\Delta G^\ominus = 237 \text{ kJ/mol}\)[/tex]
- For reaction ii: [tex]\(\Delta G^\ominus = 184.1 \text{ kJ/mol}\)[/tex]