CO(g) + H2O(g) → CO2(g) + H2(g)
From the reaction above, calculate the standard enthalpies of formation if CO2(g), H2O(g) and CO(g) in kJ mol-1 are -394, -242, and -110 respectively.
- -262 kJ mol-1
-
-42 kJ mol-1
- +42 kJ mol-1
- +262 kJ mol-1
The correct answer is: B
Explanation
CO + H2O → CO2 + H2 DH = - 394 kJ mol-1
The equation for the standard enthalpy change of formation is
ΔHreaction = ΔH(products) − ΔH(reactants)
= - 394 - ( - 242 - 110)
= - 394 - ( - 352)
= - 394 + 352
= - 42 kJ mol-1