JAMB - Chemistry (2003 - No. 27)

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

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

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