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 given reaction is: \( 2A_2(g) + B_2(g) \xrightarrow{T(K)} 2A_2B(g) + 600 \text{ kJ} \). This indicates that the reaction releases 600 kJ of energy. We need to calculate the standard enthalpy of formation \((\Delta_f H^\circ)\) of \(A_2B(g)\).

First, understand that the enthalpy change for the reaction, \(\Delta H\), is given as \(-600 \text{ kJ}\) (negative because it is exothermic). The standard enthalpy change of the reaction is related to the standard enthalpy of formation of products and reactants by the formula:

\(\Delta H = \sum(\Delta_f H^\circ \text{ of products}) - \sum(\Delta_f H^\circ \text{ of reactants})\)

For the product \(2A_2B(g)\), the enthalpy can be calculated as \(2 \cdot \Delta_f H^\circ (A_2B)\). For the reactants, since they are in their standard states, their standard enthalpies of formation are zero:

\(\Delta H = 2 \cdot \Delta_f H^\circ (A_2B) - [2 \times 0 + 0]\)

Thus the equation simplifies to:

\(-600 = 2 \cdot \Delta_f H^\circ (A_2B)\)

Solving for \(\Delta_f H^\circ (A_2B)\):

\(\Delta_f H^\circ (A_2B) = \frac{-600}{2} = -300 \text{ kJ mol}^{-1}\)

Therefore, the standard enthalpy of formation of \(A_2B(g)\) is \(-300 \text{ kJ mol}^{-1}\).

Answer 2

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