If for a certain reaction `triangle_rH` is 30KJ `mol^(-1)` at 450K, the value of `triangle_rS` (in `JK^(-1)mol^(-1))` for which the same reaction will be spontaneous at the same temperature is

Options

A. 70
B. `-33`
C. 33
D. `-70`

Answer

To determine the value of ΔS (entropy change) for which the reaction will be spontaneous at a temperature of 450 K, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

For a reaction to be spontaneous, ΔG must be negative:

\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]

This can be rearranged to find the condition for spontaneity:

\[ \Delta H < T \Delta S \]

From this, we can express ΔS in terms of ΔH and T:

\[ \Delta S > \frac{\Delta H}{T} \]

Given:
- ΔH = 30 kJ/mol = 30,000 J/mol (since 1 kJ = 1000 J)
- T = 450 K

Now, substituting the values into the equation:

\[ \Delta S > \frac{30,000 \text{ J/mol}}{450 \text{ K}} \]

Calculating the right side:

\[ \Delta S > \frac{30,000}{450} \text{ J/K mol} \]

\[ \Delta S > 66.67 \text{ J/K mol} \]

Thus, for the reaction to be spontaneous at 450 K, the value of ΔS must be greater than 66.67 J/K mol.

### Final Answer:
The value of ΔS for which the reaction will be spontaneous at 450 K is greater than 66.67 J/K mol.