When enthalpy and entropy change for a chemical reaction are `-2.5 xx10^(3)` cals and `7.4` cals `K^(-1)` respectively. Predict that reaction at 298 K is

Options

A. Spontaneous
B. Reversible
C. Irreversible
D. Non-spontaneous

Answer

To solve the problem, we need to determine whether the given chemical reaction is spontaneous at 298 K using the provided values for enthalpy change (ΔH) and entropy change (ΔS).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Enthalpy change (ΔH) = -2.5 x 10^3 cal
- Entropy change (ΔS) = 7.4 cal K^(-1)
- Temperature (T) = 298 K

2. **Use the Gibbs Free Energy Equation:**
The Gibbs free energy change (ΔG) can be calculated using the formula:
\[ \Delta G = \Delta H - T \Delta S \]

3. **Substitute the Values into the Equation:**
- First, convert ΔH into a consistent unit if necessary (it is already in calories).
- Substitute the values into the equation:
\[ \Delta G = (-2.5 \times 10^3 \text{ cal}) - (298 \text{ K}) \times (7.4 \text{ cal K}^{-1}) \]

4. **Calculate TΔS:**
- Calculate \( T \Delta S \):
\[ T \Delta S = 298 \times 7.4 = 2205.2 \text{ cal} \]

5. **Calculate ΔG:**
- Now substitute this value back into the ΔG equation:
\[ \Delta G = -2500 \text{ cal} - 2205.2 \text{ cal} = -4705.2 \text{ cal} \]

6. **Interpret the Result:**
- Since ΔG is negative (\(-4705.2 \text{ cal}\)), this indicates that the reaction is spontaneous at 298 K.

### Conclusion:
The reaction is spontaneous at 298 K because the Gibbs free energy change (ΔG) is negative.

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