Question

For a certain reaction, \(\Delta H^0\) is \(-345\) kJ and \(\Delta S^0\) is \(-123 \text{JK}^{-1}\). At what temperature the change over from spontaneous to nonspontaneous will occur?

• A. \(1052\) K
• B. \(1956\) K
• C. \(2568\) K
• D. \(2805\) K

Hint:

At the changeover temperature, delta G is zero, so T = delta H / delta S.

Answer 1

The Correct Option is D

Step 1: Condition
The changeover means $\Delta G = 0$, so $\Delta H = T\Delta S$.

Step 2: Calculate
$T = \dfrac{345 \times 10^{3}}{123} = 2804.9$ K (the signs cancel because both are negative).

Step 3: Check direction
Below this T, $|\Delta H|$ beats $|T\Delta S|$ and the reaction goes forward. Above it, it reverses. Answer: option (D).

Final Answer:
Option (D), 2805 K. \[ \boxed{2805 \text{ K}} \]