Question

Observe the following reaction: \[ 2A_2(g) + B_2(g) \xrightarrow{T(K)} 2A_2B(g) + 600 \text{ kJ} \] The standard enthalpy of formation \( (\Delta_f H^\circ) \) of \( A_2B(g) \) is:

• A. \( 600 \text{ kJ mol}^{-1} \)
• B. \( 300 \text{ kJ mol}^{-1} \)
• C. \( -300 \text{ kJ mol}^{-1} \)
• D. \( -600 \text{ kJ mol}^{-1} \)

Hint:

To find the standard enthalpy of formation per mole, divide the total enthalpy change by the number of moles of product formed. For an exothermic reaction, the enthalpy change is negative.

Answer 1

The Correct Option is C

The reaction provided is:

2A₂(g) + B₂(g) → 2A₂B(g) + 600 kJ

This is a formation reaction of A₂B from its elements A₂ and B₂ in their gaseous states. The standard enthalpy of formation (ΔfH°) is defined as the enthalpy change when 1 mole of a compound is formed from its elements in their standard states.

In this reaction, the enthalpy change is 600 kJ for the formation of 2 moles of A₂B. 
Therefore, the enthalpy of formation (ΔfH°) for 1 mole of A₂B is half of -600 kJ, which equals -300 kJ/mol.

The correct standard enthalpy of formation (ΔfH°) of A₂B(g) is: - 300 kJ mol^-1