2.4 g coal is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure. The temperature of the calorimeter rises from 298 K to 300 K. The enthalpy change during the combustion of coal is –x kJ mol^–1. The value of x is ______.(Nearest integer)
(Given : Heat capacity of bomb calorimeter is 20.0 kJ K^–1. Assume coal to be pure carbon)

Question

The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol$^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions A$^{2+}$ and B$^{4+}$ respectively is:

Answer 1

First, calculate the heat absorbed by the bomb calorimeter using the formula: $ q = C \Delta T $, where $ C $ is the heat capacity and $ \Delta T $ is the temperature change. Here, C = 20.0 kJ K^–1 and $ \Delta T $ = 300 K – 298 K = 2 K. Thus, $ q = 20.0 \times 2 = 40.0 $ kJ.
Next, we relate this heat to the combustion of coal, assuming pure carbon: Since the calorimeter absorbs the heat, it correlates directly to the combustion heat released.
Calculate moles of carbon in 2.4 g coal. Molar mass of carbon = 12.0 g mol^–1. Moles = $ \frac{2.4}{12} = 0.2 $ mol.
The enthalpy change ($ \Delta H $) per mole of carbon (coal) combustion is given as $-x$ kJ mol^–1. For 0.2 mol, the absorbed heat: $0.2 \times (-x) = -40.0$ kJ. Hence, $-0.2x = -40$.
Solve for $ x $: $ x = 200.0 $ kJ mol^–1.
Finally, confirm the obtained result $ x = 200 $ is the nearest integer, which falls within the specified range of 200,200.

Correct Answer: 200

Solution and Explanation

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