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The enthalpy change for two reactions are given by the equations 2Cr(s) +3/2O2 (g)--->Cr2O3(s) ∆H -1130kJC(s) +1/2O2 (g)--->CO(g) ∆H =-110kJWhat is enthalpy change in kJ for the reaction 3C(s)+Cr2O3 (s) --->2Cr (s) +3CO (g) is....
2Cr (s) +3/2 O2 → Cr2O3 ∆H -1130kJC(s) +1/2O2 (g)--->CO(g) ∆H =-110kJ multiply eq 2 by 3 and 1 by -1 Cr2O3 → 2Cr (s) +3/2 O2 ∆H= +1130kJ3C(s) +3/2O2 (g)---> 3CO(g) ∆H =-330kJ Add both reactions ∆H = 1130-330= 800kJ/mol So 3C(s)+Cr2O3 (s) --->2Cr (s) +3CO
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