Question

When 0.36 g of glucose was burned in a bomb calorimeter (Heat capacity $600J{K^{ - 1}}$) the temperature rise by 10 K. Calculate the standard molar enthalpy of combustion ($MJ/mole$).

Answer 1

Hint: Molar enthalpy of combustion (${\Delta _c}H$) or molar heat of combustion is the amount of heat released on complete combustion of 1 mole of the substance in the presence of oxygen at constant pressure. By convention, the value of ${\Delta _c}H$ is given as a positive value. To measure the value of molar enthalpy of combustion, a calorimeter is used.

Formula used: ${\Delta _c}H = \dfrac{{C \times \Delta T}}{n}$
${\Delta _c}H = $ Molar enthalpy of combustion
$C = $ Heat capacity
$\Delta T = $ Rise in temperature
$n = $ Number of moles of glucose burnt

Complete step by step answer:
A bomb calorimeter is used for determining and measuring the molar heat of combustion for a given chemical reaction. This type of calorimeter is a constant volume calorimeter. The term bomb is used for the fixed amount of reactant placed inside the calorimeter in a sealed condition.
Now, we have to calculate the standard molar enthalpy of combustion for combustion of glucose. Let us first write the combustion reaction for glucose.
${C_6}{H_6}{O_{12}}_{(s)} + {O_2}_{(g)} \to 6C{O_{2(g)}} + 6{H_2}{O_{(l)}}$
Let us now note down the quantities given:
Mass of glucose (m) = 0.36g
Heat capacity (C) = $600J{K^{ - 1}}$
Rise in temperature = $\Delta T$= 10K
Step 1: The first step is to calculate the number of moles of glucose burnt.
The molar mass of the compound can be calculated by summing the relative atomic masses of atoms in a molecule of that compound.
Glucose is composed of carbon, hydrogen and oxygen.
The atomic masses of carbon, hydrogen and oxygen are 12, 1 and 16 respectively.
Molar mass of ${C_6}{H_6}{O_{12}}$= $6 \times 12 + 6 \times 1 + 12 \times 16$
Molar mass of ${C_6}{H_6}{O_{12}}= 180$
The molar mass of ${C_6}{H_6}{O_{12}}$ is 180 g.
The formula for calculate moles of ${C_6}{H_6}{O_{12}}$ burnt is:
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}glucose(n) = \dfrac{{mass{\text{ }}of{\text{ }}glucose burnt}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}glucose}}$
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}glucose(n) = \dfrac{{0.36}}{{180}}$
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}glucose(n) = 0.002moles$
The number of moles of glucose burnt is 0.002moles.
Step 2: Now, we know the number of moles of glucose burnt, heat capacity and rise in temperature, we can substitute the values in the given formula and determine the molar enthalpy of combustion for the combustion of glucose.
${\Delta _c}H = \dfrac{{C \times \Delta T}}{n}$
On substituting the respective values in the above formula, we get,
${\Delta _c}H = \dfrac{{600J{K^{ - 1}} \times 10K}}{{0.002moles}}$
$\therefore {\Delta _c}H = 3 \times {10^6}J/mol = 3MJ/mol$
The standard molar enthalpy of combustion of 0.36 of is $3 \times {10^6}J/mol$.
Also note that, the molar enthalpy change for the chemical reaction is negative due to the fact that combustion reaction is always exothermic in nature. Hence, the molar enthalpy of combustion for combustion of glucose will be $ - 3 \times {10^6}J/mol$.

Additional information:
Heat capacity of a substance refers to the amount of heat released or absorbed on changing temperature through 1 degree Celsius. It is an extensive property of a substance. Heat capacity of a substance depends upon both the mass and composition of that substance.
The quantity of energy absorbs or evolves when the temperature of the substance raised by 1 degree Celsius is termed as the molar heat capacity of that substance.

Note: The standard molar enthalpy of combustion can be calculated in an alternate way as shown below.
Step 1: Calculate the number of moles of glucose molecule burnt.
$number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}glucose(n) = 0.002moles$
Step 2: Calculate the energy change for the reaction.
$C = \dfrac{q}{{\Delta T}}$
Here, C is heat capacity, q is energy change and $\Delta T$ is change in temperature.
On rearranging, $q = C \times \Delta T$
On substituting respective values, we get,
$q = 600 \times 10$
$\therefore q = 6KJ$
Step 3: Now, let us calculate the heat energy released per mole.
Heat released per mole ($\Delta U) =\dfrac{q}{n}$
Put the values in above formula,
$\Delta U = \dfrac{6}{{0.002}} = - 3MJ$
$\Delta H = \Delta U + \Delta {n_g}RT$
Here, $\Delta {n_g} = 0$
Therefore, $\Delta H = \Delta U$
$\therefore \Delta H = - 3MJ/mol$