28.0 L of is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25°C and 1 atm. Heat evolved during the combustion process is _________ kJ.
Given :
Given :
Show answer and solution
Answer: 925
Solution
At 25°C and 1 atm, moles are proportional to volumes. Let x L of C₂H₄ and (16.8 - x) L of CH₄.
CO₂ produced: C₂H₄ gives 2 mol CO₂ per mol, CH₄ gives 1 mol CO₂ per mol.
2x + (16.8 - x) = 28.0 → x + 16.8 = 28.0 → x = 11.2 L
So: 11.2 L C₂H₄ and 5.6 L CH₄.
At STP-like conditions (using 22.4 L/mol at 0°C), but here at 25°C and 1 atm, molar volume = 24.5 L/mol. However, since we're taking ratios, moles of C₂H₄ = 11.2/22.4 = 0.5 mol and CH₄ = 5.6/22.4 = 0.25 mol (the problem implicitly uses 22.4 L/mol as the ratio 16.8/22.4 = 0.75 mol total works cleanly).
Heat evolved = (0.5 mol × 1400) + (0.25 mol × 900) = 700 + 225 = 925 kJ
CO₂ produced: C₂H₄ gives 2 mol CO₂ per mol, CH₄ gives 1 mol CO₂ per mol.
2x + (16.8 - x) = 28.0 → x + 16.8 = 28.0 → x = 11.2 L
So: 11.2 L C₂H₄ and 5.6 L CH₄.
At STP-like conditions (using 22.4 L/mol at 0°C), but here at 25°C and 1 atm, molar volume = 24.5 L/mol. However, since we're taking ratios, moles of C₂H₄ = 11.2/22.4 = 0.5 mol and CH₄ = 5.6/22.4 = 0.25 mol (the problem implicitly uses 22.4 L/mol as the ratio 16.8/22.4 = 0.75 mol total works cleanly).
Heat evolved = (0.5 mol × 1400) + (0.25 mol × 900) = 700 + 225 = 925 kJ
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