Question
Chemistry Question on Enthalpy change
Given ΔHf∘ for CO2(g),CO(g) and H2O(g) are −393.5,−110.5 and −241.8kJmol−1, respectively. The ΔHf∘ [in kJmol−1 ] for the reaction CO2(g)+H2(g)⟶CO(g)+H2O(g) is
A
524.1
B
-262.5
C
-41.7
D
41.2
Answer
41.2
Explanation
Solution
CO2(g)+H2(g)⟶CO(g)+H2O(g)
ΔHf∘=(ΔH)products −(ΔH)reactants
Hence, ΔHf∘=[ΔHfCO+ΔHfH2O]−[ΔHfCO2]…(i)
Given, ΔHfCO2(s)∘=−393.5kJmol−1
ΔHfCO∘=−110.5kJmol−1
ΔHfH2O∘=−241.8kJmol−1
Put there value in equation
ΔHf∘=−110.5−241.8+393.5=41.2