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A bomb calorimeter contains 500 \mathrm{~g} of water. The heat capacity of the calorimeter system is 2000 \mathrm{~J} /{ }^{\circ} \mathrm{C}. A sample of methane \left(\mathrm{CH}_4\right) weighing 2.5 \mathrm{~g} is burned in the bomb calorimeter. The initial and final temperatures of the calorimeter system are \mathrm{25^{\circ} \mathrm{C}\: and \: 30^{\circ} \mathrm{C}}, respectively. Calculate the heat released during the combustion of methane and the corresponding enthalpy change of the reaction.

\text { (Molar mass of } C H_4=16 \mathrm{~g} / \mathrm{mol}, \mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \text { ) }

Option: 1

-3401 \mathrm{~kJ} / \mathrm{mol}
 


Option: 2

-4401 \mathrm{~kJ} / \mathrm{mol}
 


Option: 3

-2401 \mathrm{~kJ} / \mathrm{mol}
 


Option: 4

-1401 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

Given,
Mass of water = 500 g
The heat capacity of the calorimeter system = 2000 J/°C
Mass of methane burned = 2.5 g
The initial temperature of the calorimeter system = 25°C
The final temperature of the calorimeter system = 30°C
Molar mass of  \mathrm{C H_4=16 \mathrm{~g} / \mathrm{mol}}
R = 8.31 J/mol·K

First, we can calculate the heat released during the combustion of methane using the formula:

\mathrm{Q=\Delta E=C\left(T_2-T_1\right)}

Where Q is the heat released, \Delta E is the internal energy change, C is the heat capacity of the calorimeter system, and T1 and T2 are the initial and final temperatures, respectively.
\begin{aligned} & Q=2000 \mathrm{~J} /{ }^{\circ} \mathrm{C} \times\left(30^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right) \\ & Q=10,000 \mathrm{~J} \end{aligned}

Next, we can calculate the moles  of methane burned using the formula:

moles = mass/molar mass

\text { moles of } C H_4=2.5 \mathrm{~g} / 16 \mathrm{~g} / \mathrm{mol}

\begin{aligned} & \mathrm{\text {moles of } C H_4}= \\ & \mathrm{n_{\mathrm{CH}_4}=\frac{2.5 \mathrm{~g}}{16 \mathrm{~g} / \mathrm{mol}} }\\ & \mathrm{n}_{\mathrm{CH}_4}=0.15625 \mathrm{~mol} \end{aligned}

Since the combustion of 1 mole of methane releases, \mathrm{\Delta H=-890.4 \mathrm{~kJ} / \mathrm{mol}}, we can calculate the enthalpy change of the reaction as follows:

\begin{aligned} & \mathrm{\Delta H=\Delta E+\Delta n_g R T }\\ & \mathrm{\Delta n_g=\left(1 \mathrm{~mol} \text { of } \mathrm{CO}_2+2 \mathrm{~mol} \mathrm{~of~} \mathrm{H}_2 \mathrm{O}\right)-\left(1 \mathrm{~mol} \mathrm{~of} \mathrm{~CH}_4\right)} \\ & \mathrm{\Delta n_g=-1 \mathrm{~mol}} \\ & \Delta \mathrm{H}=-890.4 \mathrm{~kJ} / \mathrm{mol}+(-1 \mathrm{~mol}) \times 8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times\left(30^{\circ} \mathrm{C}+273.15 \mathrm{~K}\right) \\ & \Delta \mathrm{H}=-3409.57 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta \mathrm{H} \approx-3401 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Therefore, the heat released during the combustion of methane is 10,000 J and the corresponding enthalpy change of the reaction is -3401 \mathrm{~kJ} / \mathrm{mol}

Posted by

Ritika Kankaria

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