Question

a student runs two experiments with a constant-volume \bomb\ calorimeter containing 1500. g of water (see sketch at right).first, a 8.000 g tablet of benzoic acid ($\text{c}_{6}\text{h}_{5}\text{co}_{2}\text{h}$) is put into the \bomb\ and burned completely in an excess of oxygen. (benzoic acid is known to have a heat of combustion of 26.454 kj/g.) the temperature of the water is observed to rise from 13.00 °c to 44.35 °c over a time of 10.7 minutes.next, 5.200 g of ethane ($\text{c}_{2}\text{h}_{6}$) are put into the \bomb\ and similarly completely burned in an excess of oxygen. this time the temperature of the water rises from 13.00 °c to 46.31 °c.use this information, and any other information you need from the aleks data resource, to answer the questions below about this reaction:$ce{2c_{2}h_{6}(g) + 7o_{2}(g) → 4co_{2}(g) + 6h_{2}o(g)}$be sure any of your answers that are calculated from measured data are rounded to the correct number of significant digits.note for advanced students: its possible the student did not do these experiments sufficiently carefully, and the values you calculate may not exactly match published values for this reaction.is this reaction exothermic, endothermic, or neither?○ exothermic○ endothermic○ neitherif you said the reaction was exothermic or endothermic, calculate the amount of heat that was released or absorbed by the reaction in the second experiment.$square$ kjcalculate the reaction enthalpy $delta h_{\text{rxn}}$ per mole of $\text{c}_{2}\text{h}_{6}$.$square$ $\frac{\text{kj}}{\text{mol}}$

Explanation:

Step1: Calculate heat from benzoic acid

First, find the heat released by burning benzoic acid, which equals the heat absorbed by the calorimeter system.
$q_{benzoic} = m \times \Delta H_{combustion} = 8.000\ \text{g} \times 26.454\ \text{kJ/g} = 211.632\ \text{kJ}$

Step2: Find calorimeter constant $C_{cal}$

Use the temperature change of water to calculate $C_{cal}$, where $\Delta T_1 = 44.35\ ^\circ\text{C} - 13.00\ ^\circ\text{C} = 31.35\ ^\circ\text{C}$.
$C_{cal} = \frac{q_{benzoic}}{\Delta T_1} = \frac{211.632\ \text{kJ}}{31.35\ ^\circ\text{C}} = 6.7506\ \text{kJ/}^\circ\text{C}$

Step3: Calculate heat from ethane combustion

Find the temperature change for ethane: $\Delta T_2 = 46.31\ ^\circ\text{C} - 13.00\ ^\circ\text{C} = 33.31\ ^\circ\text{C}$. The heat released by ethane equals the heat absorbed by the calorimeter.
$q_{ethane} = C_{cal} \times \Delta T_2 = 6.7506\ \text{kJ/}^\circ\text{C} \times 33.31\ ^\circ\text{C} = 224.86\ \text{kJ}$

Step4: Moles of ethane reacted

Molar mass of $\text{C}_2\text{H}_6$ is $2(12.01) + 6(1.008) = 30.068\ \text{g/mol}$.
$n_{\text{C}_2\text{H}_6} = \frac{5.200\ \text{g}}{30.068\ \text{g/mol}} = 0.1729\ \text{mol}$

Step5: Calculate $\Delta H_{\text{rxn}}$ per mole of $\text{C}_2\text{H}_6$

Since heat is released, $\Delta H$ is negative.
$\Delta H_{\text{rxn}} = -\frac{q_{ethane}}{n_{\text{C}_2\text{H}_6}} = -\frac{224.86\ \text{kJ}}{0.1729\ \text{mol}} = -1301\ \text{kJ/mol}$

Answer:

Is this reaction exothermic, endothermic, or neither?
exothermic

Heat released in the second experiment:
225 kJ

Reaction enthalpy $\Delta H_{\text{rxn}}$ per mole of $\text{C}_2\text{H}_6$:
$-1301\ \frac{\text{kJ}}{\text{mol}}$