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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....

User , 7 Years ago
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Shailendra Kumar Sharma

Last Activity: 7 Years ago

2Cr (s) +3/2 O2 → Cr2O3 ​ ∆H -1130kJ
C(s) +1/2O2 (g)--->CO(g) ∆H =-110kJ 
multiply eq 2 by 3 and 1 by -1 
 Cr2O3​ → ​2Cr (s) +3/2 O2​ ​∆H= +1130kJ
3C(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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