Question

When 225mg of anthracene, C14H10(s), was burned in a bomb calorimeter the temperature rose by 1.75K. Calculate the calorimeter constant. By how much will the temperature rise when 125mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? (ΔcH<(C14H10,s)=–7061 kJ mol−1.)

Answer 1

Answer:

Calorimeter constant: $0.0227\frac{kJ}{mg*K}$

Temperature raise for phenol: $\Delta T=1.43K$

Explanation:

Hello,

In this case, since the combustion of anthracene was carried out inside the calorimeter, the released heat equals the negative of its enthalpy of combustion and the reacted moles of anthracene whose molar mass is 178 g/mol:

$Q=n*-\Delta _CH=225mg*\frac{1g}{1000mg} *\frac{1mol}{178g} -(-7061\frac{kJ}{mol} )=8.93kJ$

Thus, we can compute the calorimeter constant by knowing the temperature increase and the combusted mass:

$Q=mC\Delta T\\\\C=\frac{Q}{m\Delta T} =\frac{8.93kJ}{225mg*1.75K}$

$C=0.0227\frac{kJ}{mg*K}$

Next, for 125 mg of phenol whose molar mass is 96 g/mol and enthalpy of combustion is -3051 kJ/mol we compute the temperature rise for the same conditions by firstly computing the involved heat:

$Q=n*-\Delta _CH=125mg*\frac{1g}{1000mg} *\frac{1mol}{94g} -(-3051\frac{kJ}{mol} )=4.06kJ$

Then:

$Q=mC\Delta T\\\\\Delta T=\frac{Q}{mC} =\frac{4.06kJ}{125mg*0.0227\frac{kJ}{mg*K} }\\\\\\\Delta T=1.43K$

Regards.