Question

A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature rises from 25.0 to 62.3°C, determine the heat capacity of the calorimeter. The molar mass of ethanol is 46.07 g/mol. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) ΔH°rxn = -1235 kJ

Answer 1

Answer: The heat capacity of calorimeter is $15.66J/^oC$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol$

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

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

where,

$q$ = amount of heat released = ?

n = number of moles = 0.473 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction  = -1235 kJ/mol = $-1235\times 10^3J/mol$     (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

$-1235\times 10^3J/mol=\frac{q}{0.473mol}\\\\q=(-1235\times 10^3J/mol\times 0.473mol)=-584.16\times 10^3J$

To calculate the heat capacity of calorimeter, we use the equation:

$q=c\Delta T$

where,

q = heat absorbed by the calorimeter = $584.16\times 10^3J$

c = heat capacity = ?

$\Delta T$ = change in temperature = $T_2-T_1=62.3^oC-25^oC=37.3^oC$

Putting values in above equation, we get:

$584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC$

Hence, the heat capacity of calorimeter is $15.66J/^oC$