Answer:
The ΔH for the reaction is -456.5 KJ
Explanation:
Here we want to determine ΔH for the reaction;
Mathematically;
ΔH = ΔH(product) - ΔH(reactant)
In the case of the first reaction;
ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)
From the other reactions, we can get the respective ΔH for the individual molecule in the reaction
In second reaction;
Kindly note that for elements, molecule of gases, ΔH = 0
What this means is that throughout the solution;
ΔH(Ca) = 0 KJ
ΔH(O2) = 0 KJ
ΔH(C) = 0 KJ
Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone
So in the second reaction
ΔH = 2ΔH(CaCO3)
Thus;
-2414/2 = ΔH(CaCO3)
ΔH(CaCO3) = -1,207 KJ
Moving to the third reaction, we have;
ΔH = ΔH(CO2)
Hence ΔH(CO2) = -393.5 KJ
For the last reaction;
ΔH = ΔH(CaO)
Hence ΔH(CaO) = -1270 KJ
Going back to equation *
ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)
Using the values of the ΔH of the respective molecules given above,
ΔH = -1270 + (-393.5) - (-1207)
ΔH = -456.5 KJ
