Question
Many reactions that have large, negative values of $\Delta G^{\circ}$ are not actually observed to happen at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$. Why?
Step 1
It is the Gibbs free energy change under standard conditions (25°C and 1 atm). A negative value of $\Delta G^{\circ}$ indicates that the reaction is spontaneous and the equilibrium lies far towards the product side. Show more…
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In your own words, explain why (a) $\Delta S^{\circ}$ is negative for a reaction in which the number of moles of gas decreases. (b) we take $\Delta S^{\circ}$ to be independent of $T$, even though entropy increases with $T$. (c) a solid has lower entropy than its corresponding liquid.
In your own words, explain why (a) $\Delta S^{\circ}$ is negative for a reaction in which the number of moles of gas decreases. (b) we take $\Delta S^{\circ}$ to be independent of $T,$ even though entropy increases with $T$. (c) a solid has lower entropy than its corresponding liquid.
Calculate the $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$ for the preceding reaction. In your own words, explain why a reaction that combines two molecules into one should have a large negative entropy change.
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