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: Understanding the Concept
A reaction is spontaneous when \(\Delta G < 0\). Since \(\Delta G = \Delta H - T\Delta S\), the switch between spontaneous and nonspontaneous happens where \(\Delta G = 0\).

Step 2: Key Formula or Approach
\[ T = \frac{\Delta H}{\Delta S} \]

Step 3: Detailed Explanation
Convert units first. \(\Delta H = -345\) kJ = \(-345000\) J, and \(\Delta S = -123\) J/K.
\[ T = \frac{-345000}{-123} = 2804.9 \approx 2805 \text{ K} \]
Both \(\Delta H\) and \(\Delta S\) are negative. Below 2805 K the \(-T\Delta S\) term is small, so \(\Delta G\) is negative and the reaction is spontaneous. Above it, \(\Delta G\) turns positive. Leaving \(\Delta H\) in kJ would give a wrong value near 2.8 K.

Final Answer:
The changeover occurs at 2805 K, option (D). \[ \boxed{2805 \text{ K}} \]