Question

The enthalpy of combustion of benzoic acid (CH-COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters, its value has been accurately determined to be -3226.7 kJ mol! (a) When 0.9862 g of benzoic acid was oxidized, the temperature rose from 21.84°C to 25.67°C. What is the heat capacity of the calorimeter? (b) In a separate experiment, 0.4654 g of glucose (C6H120) was oxidized in the same calorimeter, and the temperature rose from 21.22°C to 22.28°C. Calculate the enthalpy of combustion of glucose, the value of A U for the combustion, and the molar enthalpy of formation of glucose.

Answer 1

Given data set:

0.9862 g of benzoic acid is oxidized in a caloriemeter, then the temperature raises from 21.84 C to 25.67 C.

We have to calculate the heat capacity of the caloriemeter.

Given, enthalpy of combustion of benzoic acid is -3226.7 KJ mol-1 which means upon combustion of 1 mole i.e 122.1 g of benzoic acid 3226.7 KJ of energy is released(negative sign indicates heat is been released).

So, combustion of 0.9862 g of benzoic acid will produce:

Qrxn = _{$\frac{1 mol}{122.1 g}*(0.9862 g)*(3226.7 KJ/mol)$}

Qrxn = 26.062 KJ

Now, As the caloriemeter is an isolated system, all the heat released will be observed by the bomb and water surrounding it.

Thus, heat absorbed by bomb can be written as:

Qbomb = C*$\Delta$t

   where C is the heat capacity of the bomb

  $\Delta$t is the change in temperature.

Also, heat absorbed by water can be written as:

Qwater = m*s*$\Delta$t

   where m is the mass of water

   s is the specific heat of water

  $\Delta$t is the change in temperature.

So, we can write:

Qrxn = Qbomb + Qwater

= C*$\Delta$t + m*s*$\Delta$t ................... Equation 1

C = (Qrxn - m*s*$\Delta$t)/$\Delta$t

$\Delta$t = 25.67 C - 21.84 C

= 3.83 C

Substituting all given in Equation 1:

We have:

C = (26.062 KJ - 2250*4.18 kJ/kg K*3.83 C)/3.83

   = 1.98 KJ/g K