Question

A 0.56 g sample of liquid C6H12 was combusted completely using excess oxygen inside a bomb (constant volume) calorimeter, with the products being carbon dioxide and liquid water. The calorimeter's heat capacity is 5.36 kJ °C-1. If the temperature inside the calorimeter increased from 25.0 °C to 29.2 °C, determine ΔrH for this reaction in kJ mol-1 (with respect to C6H12) at 298 K. Do not worry about how realistic the final answer is. You have 5 attempts at this question. TIP: To report an answer in scientific notation, enter it using the format "2.3E4", which means "2.3 x 104" (without the quotation marks)

Answer 1

-3.4 × 10³ kJ/mol

Step-by-step explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

`Qcal = c_(cal) * \Delta T`

where,

`Qcal`: heat absorbed by the calorimeter

`c_(cal)`: heat capacity of the calorimeter

`\Delta T`: change in the temperature

`Qcal = (5.36kJ)/(\° C) * (29.2\° C - 25.0 \° C ) = 22.5 kJ`

Step 2: Calculate the heat released by the combustion

According to the law of conservation of energy, the sum of the heat absorbed by the calorimeter and the heat released by the combustion is zero.

`Q_(cal) + Q_(comb) = 0\\Q_(comb) = -Q_(cal) = -22.5 kJ`

Step 3: Calculate the standard enthalpy of the combustion

22.5 kJ are released when 0.56 g of C₆H₁₂ (MW 84.16) is combusted. The standard enthalpy of the combustion (ΔH°c) is:

`(-22.5kJ)/(0.56g) * (84.16g)/(mol) = -3.4 * 10^(3) kJ/mol`