Question

1. At constant volume, the heat of combustion of a particular compound, compound A, is –3409.0 kJ/mol. When 1.277 g of compound A (molar mass = 117.77 g/mol) was burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 6.891 °C. Using this data, what is the heat capacity (calorimeter constant) of the calorimeter?

2. Suppose a 3.107 g sample of a second compound, compound B, was combusted in the same calorimeter, and the temperature rose from 24.37 °C to 30.30 °C. What is the heat of combustion per gram of compound B?

Answer 1

Solution- Mass of compound A= 1.277g

Moles of compound A=Mass/molar mass=1.277/117.77 =0.01084 mol
So,
Heat released in reaction=Moles of compound*heat of combustion

                                =0.01084*3409.0 =36.95 kJ

Let C be the heat capacity of calorimeter be.

So,

C*change in temperature=Heat released in reaction

C*6.891=36.95

Calorimeter constant, C=5.36 kJ/C

…………………………………………………………………………….

Let us suppose heat of combustion be Δh
So,
Mass*Δh=Calorimeter constant*Change in temperature

3.107*Δh=5.36*(30.30-24.37)

Δh=10.23 kJ/g

= -10.23 kJ/g Answer