Question

Consider the following reaction:
2 C₆H₆+15 O₂→12 CO₂+6 H₂O \(Δ_r𝐻^0_{298}\) =−3120 kJ mol⁻¹.
A closed system initially contains 5 moles of benzene and 25 moles of oxygen under standard conditions at 298 K. The reaction was stopped when 17.5 moles of oxygen is left. The amount of heat evolved during the reaction is____kJ. 
(round off to the nearest integer)

Answer 1

Correct Answer: 1560

To determine the amount of heat evolved during the given reaction, we first need to identify the limiting reactant, as it will dictate the extent of the reaction.

Step 1: Initial Moles and Reaction Progress
We have the reaction:
2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O
Initially, there are 5 moles of C₆H₆ and 25 moles of O₂. The reaction stops with 17.5 moles of O₂ remaining.

Step 2: Determine Moles of O₂ Reacted
Moles of O₂ reacted = Initial O₂ - Remaining O₂ = 25 - 17.5 = 7.5 moles.

Step 3: Calculate Moles of C₆H₆ Reacted
From the stoichiometry (15:2 ratio),
\(\frac{7.5 \text{ moles O}_2}{15} \times 2 = 1\text{ mole C}_6\text{H}_6\).

Step 4: Calculate Heat Evolved
The given Δ_rH⁰₂₉₈ is -3120 kJ for 2 moles of C₆H₆. Thus, for 1 mole of C₆H₆:
Energy evolved = \(\frac{-3120}{2} \text{ kJ} = -1560 \text{ kJ}\).

Step 5: Confirm Value Within Range
The calculated value is -1560 kJ, which fits within the provided range. Rounding is not necessary as it matches the expected precision.

The amount of heat evolved during the reaction is 1560 kJ.