Combustion of 7.54 g of liquid benzene (C6H6) causes a temperature rise of 50.3°C in a constant-pressure calorimeter that has a heat capacity of 6.27 kJ/°C. What is ∆H for the following reaction? C6H6 (l) + O2 (g) → 6CO2 (g) + 3H2O (l)

Question

Chemistry

Rocko

21 April, 14:20

Combustion of 7.54 g of liquid benzene (C6H6) causes a temperature rise of 50.3°C in a constant-pressure calorimeter that has a heat capacity of 6.27 kJ/°C. What is ∆H for the following reaction? C6H6 (l) + O2 (g) → 6CO2 (g) + 3H2O (l)

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Cristofer Allen · Answer 1

-3268 kJ/mol

Explanation:

The calorimeter is an equipment used to measure the heat from a combustion reaction. The heat can be calculated by:

Q = C*ΔT

Where Q is the heat, C is the heat capacity of the calorimeter, and ΔT is the variation of the temperature. So:

Q = 6.27 * 50.3

Q = 315.381 kJ

Because the calorimeter is adiabatic, the heat absorbed by it must be equal to the heat released for the reaction, so the heat of the reaction is:

Q = - 315.381 kJ

The change in enthalpy (ΔH) is the heat divided by the number of moles of the limiting reactant. The combustion is commonly done with oxygen in excess, so benzene is the limiting reactant. Benze has a molar mass equal to 78.11 g/mol, so the number of moles is:

n = mass/molar mass

n = 7.54/78.11

n = 0.0965 moles

ΔH = Q/n

ΔH = - 315.381/0.0965

ΔH = - 3268 kJ/mol