## Test 2 #5

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

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### Test 2 #5

On the most recent test there was the question:

Given that $\Delta H$rxno = -42 kJ/mol and K = 1.63106 for a reaction at 25 C, calculate K at 110 C. Assume that $\Delta H$rxno and $\Delta S$rxno remain constant over this temperature range.

The equation sheet had the Van't Hoff Equation written as $ln \frac{K1}{K2} = \frac{\Delta H}{R}(\frac{1}{T1}-\frac{1}{T2})$ when it should have been $ln \frac{K1}{K2} = -\frac{\Delta H}{R}(\frac{1}{T1}-\frac{1}{T2})$. I used the second equation and got that K2 = 64 but got the question wrong.

Was there an error in grading or was I incorrect in thinking the given equation was wrong? One of my friends used the given equation and got 0.04 which was the correct answer.

Chem_Mod
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### Re: Test 2 #5

The given equation is correct. It is the same as $ln\frac{K_{2}}{K_{1}}=-\frac{\Delta H}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$, just that the negative sign was distributed inside the parentheses. Another way to more qualitatively assess the validity of your answer was to recall from the equilibrium concepts you learned at the beginning of the quarter. If you have an exothermic reaction (negative change in enthalpy), increasing the temperature will shift the equilibrium towards reactants by decreasing K.