Question

The heat evolved during the combustion of 112 liter of water gas at STP (mixture of equal volume of ${H_2}$ and $CO$ is:
Given:
${H_2}(g) + 1/2{O_2}(g) \to {H_2}O(g);\Delta H = - 241.8kJ$
$CO(g) + 1/2{O_2}(g) \to C{O_2};\Delta H = - 283kJ$
A.241.8 kJ
B.283 kJ
C.1312 kJ
D.1586 kJ

Answer 1

Hint: The heat of combustion depends on the number of moles of reactants and products. That is, on increasing the number of moles of reactants by two times, the heat of combustion is also doubled. Also, at STP 1 mole of a gas occupies 22.4 L of volume.

Complete step by step answer:
As given in the question, the combustion reactions of ${H_2}$ and $CO$ are:
${H_2}(g) + 1/2{O_2}(g) \to {H_2}O(g);\Delta H = - 241.8kJ$
$CO(g) + 1/2{O_2}(g) \to C{O_2};\Delta H = - 283kJ$
We can see that the energies evolved in these reactions are due to the combustion of 1 moles of ${H_2}$ and $CO$ each.
Water gas is the mixture of equal volumes of ${H_2}$ and $CO$. So, the above energies combined will give the heat of combustion of water gas.
For water gas, $\Delta {H_{combustion}} = - 241.8 + ( - 283)kJ$
$ \Rightarrow \Delta {H_{combustion}} = - 524.8kJ$
At STP, the volume of 1 mole of gas is 22.4 L. As, equal volume of both ${H_2}$ and $CO$ are present so the total volume of water gas is $22.4L + 22.4L = 44.8L$.
So, 112 L of water gas $ = \dfrac{{112}}{{44.8}}$ moles
$ \Rightarrow 2.5$ moles
Therefore, heat of combustion of 2.5 moles of water gas
$ = 2.5 \times ( - 524.8)kJ$
$ \Rightarrow 1312kJ$

Hence option C is correct.

Note:
In some cases, where a set of reactions are given and the reactions need to be rearranged in order to get the desired product, then the same operations are done with the heat of combustion of the respective reactions. That is, if a reaction is reversed then the value of heat of combustion $(\Delta H)$ remains same but the sign is reversed; if a reaction is halved, then the $\Delta H$ value is also halved. Similarly, if two reactions are added the $\Delta H$ values are also added and if they are subtracted the $\Delta H$ values are also subtracted.