Question

A reaction has an equilibrium constant of 7.9×10³ at 298 K. At 713 K, the equilibrium constant is 0.77.

Do you predict that the enthalpy of the reaction is positive or negative? Why?

a. Positive, because endothermic reactions shift toward products at higher temperatures
b. Negative, because exothermic reactions shift toward reactants at higher temperatures
c. Positive, because endothermic reactions shift toward reactants at higher temperatures
d. Negative, because exothermic reactions shift toward products at higher temperatures

Answer 1

option b is correct

Step-by-step explanation:

Taking the Van't Hoff equation

d ln Keq / dT = ΔH/(RT²)

then

Keq increases with increasing temperature (d ln Keq / dT>0) when ΔH>0 and decreases with increasing temperature (d ln Keq / dT<0) when ΔH<0

if Keq decreases (reactions shift toward reactants) from 7.9×10³ to 0.77 when temperature increases from 298 K to 713 K (d ln Keq / dT<0) → ΔH<0 ( exothermic reaction)

therefore option b is correct