Question

Question #1adf2

Answer 1

-5660 kJ/mol

Explanation:

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`

`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