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Answer 1 · 2014-04-29T23:15:30

-5660 kJ/mol

The molar heat of combustion of sucrose is -5660 kJ/mol.

This is a standard calorimetry problem. The basic principle is that all the heats involved must add up to zero.

Heat from combustion + Heat to warm water = 0

$q_{1} + q_{2} = 0$ $q_1 +q_2 = 0$

$nΔH^o + mcΔT$ = 0

$n$ = 2.50 g C₁₂H₂₂O₁₁ × $(1" mol C₁₂H₂₂O₁₁")/(342.3" g C₁₂H₂₂O₁₁")$ = 7.30 × 10⁻³ mol C₁₂H₂₂O₁₁

$ΔT$ = (25.01 – 20.50) °C = 4.51 °C = 4.51 K

7.30 × 10⁻³ mol × $ΔH^o$ + 2190 g × 4.184 J·K⁻¹g⁻¹ × 4.51 K = 0

7.30 × 10⁻³ mol × $ΔH^o$ + 41 300 J = 0

$ΔH^o = -(41 300" J")/(7.30 × 10⁻³" mol")$ = -5 660 000 J/mol = -5660 kJ/mol