If ΔH < 0 and ΔS > 0, what can be said about ΔG at all temperatures?

If ΔH < 0 and ΔS > 0, what can be said about ΔG at all temperatures?

• A. ΔG is positive at high temperatures.
• B. ΔG is negative at all temperatures.
• C. ΔG is zero at some temperatures.
• D. ΔG is undefined.

Answer: (B)

The key idea is how Gibbs free energy relates to enthalpy and entropy: ΔG = ΔH - TΔS. When ΔH is negative and ΔS is positive, both terms push ΔG downward. Since ΔS > 0, the term TΔS is positive, so subtracting it makes ΔG even more negative. With ΔH < 0, at temperature zero you have ΔG = ΔH, which is negative; for any positive temperature, ΔG becomes strictly more negative. Therefore ΔG is negative at all temperatures. This means the reaction is spontaneous at all temperatures because a negative ΔG indicates spontaneity. The other possibilities don’t fit: high temperatures don’t flip the sign to positive when entropy favors the process, zero would require a negative temperature or opposing signs, and undefined isn’t the case since the equation yields a definite value for all real temperatures.

The key idea is how Gibbs free energy relates to enthalpy and entropy: ΔG = ΔH - TΔS. When ΔH is negative and ΔS is positive, both terms push ΔG downward. Since ΔS > 0, the term TΔS is positive, so subtracting it makes ΔG even more negative. With ΔH < 0, at temperature zero you have ΔG = ΔH, which is negative; for any positive temperature, ΔG becomes strictly more negative. Therefore ΔG is negative at all temperatures.

This means the reaction is spontaneous at all temperatures because a negative ΔG indicates spontaneity. The other possibilities don’t fit: high temperatures don’t flip the sign to positive when entropy favors the process, zero would require a negative temperature or opposing signs, and undefined isn’t the case since the equation yields a definite value for all real temperatures.