What can be concluded about the values of ΔH and ΔS from this graph?

Engineering

Chemistry

Gibbs Free Energy Calculations and Third Law

Question

ΔH < 0, ΔS > 0

ΔH > 0, ΔS < 0

ΔH < 0, ΔS < 0

ΔH > 0, ΔS > 0

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The graph shows ln(K) vs 1/T, which is linear with a negative slope. This relates to the van't Hoff equation: $\ln (K) = - \frac{Δ H}{R} \cdot \frac{1}{T} + \frac{Δ S}{R}$ . The slope equals -ΔH/R. A negative slope means -ΔH/R is negative, so ΔH must be positive. The y-intercept is positive, which equals ΔS/R. Therefore, ΔS must also be positive.

Final answer: ΔH > 0, ΔS > 0