Chemistry 2006 JAMB Past Questions - Question 43: H2(g) + Br2(g) 2HBr(g) The above reaction is carried out at 25°C. ΔH is -72 kJ mol-1 and ΔS is - 106 J mol-1K-1, the reaction will

Correct Answer

B

Explanation

To determine whether the reaction will proceed spontaneously at a given temperature, you can use the Gibbs free energy equation:

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

Where:
- \(\Delta G\) is the change in Gibbs free energy,
- \(\Delta H\) is the change in enthalpy,
- \(\Delta S\) is the change in entropy,
- \(T\) is the temperature in Kelvin.

If \(\Delta G\) is negative, the reaction will proceed spontaneously. If \(\Delta G\) is positive, the reaction will not proceed spontaneously.

\[ \Delta G = -72 \, \text{kJ/mol} - (25 + 273.15) \times \left(-106 \, \text{J/mol} \cdot \text{K}\right) \]

\[ \Delta G = -72 \, \text{kJ/mol} + 298.15 \times 106 \, \text{J/mol} \]

\[ \Delta G = -72 \, \text{kJ/mol} + 31.598 \, \text{kJ/mol} \]

\[ \Delta G = -40.402 \, \text{kJ/mol} \]

Since \(\Delta G\) is negative, the reaction will proceed spontaneously at the given temperature (25°C). Therefore, the correct option is: **proceed spontaneously at the given temperature.**