- #1

Sean1218

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I did a graph of ln K (y) vs 1/T (x), where K is the rate constant, and T is absolute temperature. Following the lab experiment instructions, where it says the slope is -ΔH/R, I get about 34700 J for ΔH (about the same as the values they give), and very similar ΔS (about 137 for most) values as well.

However, it's an exothermic reaction, so why do both of us have positive enthalpy values? They even say that ΔH is negative "as expected", but they list a positive value. And they wouldn't have gotten the ΔS values they listed (which they say is also as expected) if they hadn't used the positive ΔH value in the ΔG = ΔH - TΔS calculations.

What's going on?