Question

A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon.

Answer 1

Answer: The enthalpy of the reaction is 269.4 kJ/mol

Explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

$q_1=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_1=675J/^oC\times 29.62^oC=19993.5J$

To calculate the heat absorbed by water, we use the equation:

$q_2=mc\Delta T$

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)^oC=29.62^oC$

Putting values in above equation, we get:

$q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[19993.5+114690.12]J=134683.62J=134.7kJ$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol$

Hence, the enthalpy of the reaction is 269.4 kJ/mol

Answer 2

Answer : The enthalpy change per mole of hydrocarbon is, 269 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $675J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water = 925 g

$\Delta T$ = change in temperature = $T_2-T_1=(53.88-24.26)=29.62^oC$

Now put all the given values in the above formula, we get:

$q=[(675J/^oC\times 29.62^oC)+(925g\times 4.18J/g^oC\times 29.62^oC)]$

$q=134519.23J=134.5kJ$

Now we have to calculate the enthalpy change per mole of hydrocarbon.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 134.5 kJ

n = moles of hydrocarbon = 0.500 mol

$\Delta H=\frac{134.5kJ}{0.500mole}=269kJ/mole$

Therefore, the enthalpy change per mole of hydrocarbon is, 269 kJ/mole