Ask Question
1 October, 11:26

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

+4
Answers (1)
  1. 1 October, 12:19
    0
    option b is correct

    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers