Question #1adf2

1 Answer

Answer:

-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