Spontaneity, Entropy, and Free Energy Question Question 6

\ce{2H2O2(aq) -> 2H2O(l) + O2(g) \hskip{2em} \Delta H^\circ = \pu{-196 kJ/mol_{rxn}}}

The decomposition of $\ce{H2O2(aq)}$ is represented by the equation above. The reaction is thermodynamically favorable. The signs of $\Delta G^\circ$ and $\Delta S^\circ$ for the reaction are which of the following?

\Delta G^\circ \gt 0

and

\Delta S^\circ \gt 0

\Delta G^\circ \lt 0

and

\Delta S^\circ \gt 0

\Delta G^\circ \gt 0

and

\Delta S^\circ \lt 0

\Delta G^\circ \lt 0

and

\Delta S^\circ \lt 0

Approach

A thermodynamically favorable reaction implies a negative $\Delta G^\circ$ value.

Since the number of molecules of products is greater than the number of molecules in the reactants, the entropy also increased, so $\Delta S^\circ \gt 0$ .