Question
Question: The heat evolved during the combustion of 112 liter of water gas at STP (mixture of equal volume of ...
The heat evolved during the combustion of 112 liter of water gas at STP (mixture of equal volume of H2 and CO is:
Given:
H2(g)+1/2O2(g)→H2O(g);ΔH=−241.8kJ
CO(g)+1/2O2(g)→CO2;ΔH=−283kJ
A.241.8 kJ
B.283 kJ
C.1312 kJ
D.1586 kJ
Solution
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 H2 and CO are:
H2(g)+1/2O2(g)→H2O(g);ΔH=−241.8kJ
CO(g)+1/2O2(g)→CO2;ΔH=−283kJ
We can see that the energies evolved in these reactions are due to the combustion of 1 moles of H2 and CO each.
Water gas is the mixture of equal volumes of H2 and CO. So, the above energies combined will give the heat of combustion of water gas.
For water gas, ΔHcombustion=−241.8+(−283)kJ
⇒ΔHcombustion=−524.8kJ
At STP, the volume of 1 mole of gas is 22.4 L. As, equal volume of both H2 and CO are present so the total volume of water gas is 22.4L+22.4L=44.8L.
So, 112 L of water gas =44.8112 moles
⇒2.5 moles
Therefore, heat of combustion of 2.5 moles of water gas
=2.5×(−524.8)kJ
⇒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 (ΔH) remains same but the sign is reversed; if a reaction is halved, then the ΔH value is also halved. Similarly, if two reactions are added the ΔH values are also added and if they are subtracted the ΔH values are also subtracted.