Question

0.16 g of methane was subjected to combustion at $27{}^\circ \text{C}$ in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by $0.5{}^\circ \text{C}$. The heat of combustion of methane at:
(I) constant volume and (II) constant pressure
(Given: The thermal capacity of the calorimeter system is $17.7\text{kJ }{{\text{K}}^{1-}}$. R = $8.314\text{ }\text{J mo}{{\text{l}}^{1-}}\text{ }{{\text{K}}^{1-}}$).
A. (I) $-885 kJ/mol$ (II) $-889.95 kJ / mol$
B. (I) $-785 kJ/mo$l (II) $-859.95 kJ / mol$
C. (I) $-587 kJ/mol$ (II) $-789.95 kJ / mol$
D. (I) $-985 kJ/mo$l (II) $-999.95 kJ / mol$

Answer 1

For this problem, we have to use a different formula for different conditions. For constant volume heat of combustion is equal to the internal energy or constant pressure, the heat of combustion is equal to the enthalpy.

Complete Step-By-Step Answer:
- In the given question we have to calculate the value of heat of combustion of the methane at constant volume and constant pressure.
- So, firstly we have to write the reaction of the combustion of the methane that is:
$\text{C}{{\text{H}}_{4(g)}}\text{ + 2}{{\text{O}}_{2(g)}}\text{ }\to \text{ C}{{\text{O}}_{2}}_{(g)}\text{ + 2}{{\text{H}}_{2}}{{\text{O}}_{(\text{l)}}}$
- So, at constant volume as we know that the heat of combustion is equal to the internal energy.
- Internal energy is defined as the net energy of the closed system which is equal to the product of heat capacity of calorimeter, rise in temperature and molar mass of the compound divided by the mass of the compound.
$\vartriangle \text{E = Internal energy}$
- Now, it is given in the question that the heat capacity of the calorimeter is $17.7\text{kJ }{{\text{K}}^{1-}}$, rise in the temperature is $0.5{}^\circ \text{C}$, the molar mass of the methane is $12\text{ + 4 }\times \text{ 1 = 16}$and mass is 0.16g.
- So, the heat of combustion will be:
$\vartriangle \text{E = }\dfrac{\text{Heat capacity }\times \text{ temperature rise }\times \text{ Molar mass}}{\text{Mass of the compound}}$
$\vartriangle \text{E = -}\dfrac{\text{17}\text{.7 }\times \text{ 0}\text{.5 }\times \text{ 16}}{0.16}\text{ = -885 kJ/mol}$
- Now, the heat of combustion at constant pressure will be equal to the enthalpy.
- So, as we know the formula of enthalpy is the addition of heat of combustion and product of moles, gas constant and temperature i.e.
$\text{Heat of combustion = }\vartriangle \text{H (enthalpy)}$
$\vartriangle \text{H = }\vartriangle \text{E + }\vartriangle \text{nRT}$ …. (1)
-Now, here n is equal to the no. of a mole of gas in the product minus no. of moles of gas in reactant i.e. 1 - 3 = -2.
- So, equation first becomes:
$\vartriangle \text{H = 885 + (-2 }\times \text{ 8}\text{.314 }\times \text{ 1}{{\text{0}}^{3-}}\times \text{ 300) = -889}\text{.9 kJ/mol}$
Therefore, option A is the correct answer.

Note: In the given problem, enthalpy is defined as the net internal energy which is present in the system and represented by $\vartriangle H$. When the value of internal energy is negative it means that the work is done by the system in the surrounding.