Question
For the reaction$$\mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g)$$the value of $\Delta G^{\circ}$ is $-374 \mathrm{~kJ} .$ Use this value and data from Appendix 4 to calculate the value of $\Delta G_{f}^{0}$ for $\mathrm{SF}_{4}(g)$.
Step 1
Step 1: We are given the reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g) $$ and we know that the standard Gibbs free energy change for this reaction, $\Delta G^{\circ}$, is $-374 \mathrm{~kJ}$. Show more…
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For the reaction $$\mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g)$$ the value of $\Delta G^{\circ}$ is $-374 \mathrm{kJ}$ . Use this value and data from Appendix 4 to calculate the value of $\Delta G_{\mathrm{f}}^{\circ}$ for $\mathrm{SF}_{4}(g) .$
For the reaction $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+4 \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+\mathrm{H}_{2}(g)$$ $\Delta S^{\circ}$ is equal to $-358 \mathrm{J} / \mathrm{K}$ . Use this value and data from Appendix 4 to calculate the value of $S^{\circ}$ for $\mathrm{CF}_{4}(g) .$
For the reaction $$ \mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g) $$ $\Delta S^{\circ}$ is equal to $-143 \mathrm{~J} / \mathrm{K}$. Use this value and data from Appendix 4 to calculate the value of $S^{\circ}$ for $\mathrm{CS}_{2}(\mathrm{~g})$.
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