Question

question 8 of 30
for a reaction, $\delta h^0 = 176\\ \text{kj/mol}$ and $\delta s^0 = 0.285\\ \text{kj/(k·mol)}$. at what temperatures is this reaction spontaneous?
\bigcirc a. $t < 617\\ \text{k}$
\bigcirc b. $t > 617\\ \text{k}$
\bigcirc c. $t < 50\\ \text{k}$
\bigcirc d. at no temperature

Explanation:

Step1: Recall the spontaneity condition

A reaction is spontaneous when $\Delta G^0 < 0$. The formula for Gibbs free energy change is $\Delta G^0=\Delta H^0 - T\Delta S^0$.

Step2: Set up the inequality for spontaneity

Substitute into the spontaneity condition: $\Delta H^0 - T\Delta S^0 < 0$. We know $\Delta H^0 = 176\space kJ/mol$ and $\Delta S^0=0.285\space kJ/(K\cdot mol)$. So, $176 - T\times0.285 < 0$.

Step3: Solve the inequality for T

Rearrange the inequality: $T\times0.285> 176$. Then, $T > \frac{176}{0.285}$. Calculate $\frac{176}{0.285}\approx617\space K$. So the reaction is spontaneous when $T > 617\space K$.

Answer:

B. \( T> 617\space K \)