Question

21. Enthalpy changes for two reactions are given by
the equation
-
Cr2O3(s), AH = -1130 kJ
-
CO(g),
AH=-110 kJ
The enthalpy change for the reaction
Cr,O, + 3C 2 Cr + 3C0 will be​

Answer 1

Explanation:

C(s) + 2 H2(g) → CH4(g)

The formation reaction for carbon dioxide (CO2) is

C(s) + O2(g) → CO2(g)

Answer 2

Answer:

800 kJ/mol.

Explanation:

For the first reaction we have 2Cr (s) +3/2 O2 -> Cr2O3 , for this the enthalpy change is given in the question as ​ ∆H -1130kJ .

Again for the second reaction C(s) +1/2O2 (g)->CO(g) the change in the enthalpy is given in the question as ∆H =-110kJ.  

multiply eq 2 by 3 and 1 by -1  

To equate the equation the yield the net result we multiply the first equation by -1 and the second by 3.

So, we will get Cr2O3​ -> ​2Cr (s) +3/2 O2 with​ ​∆H= +1130kJ

3C(s) +3/2O2 (g) -> 3CO(g) and ∆H =-330kJ .

Therefore, if we add both the equations we will get the result reaction 3C(s)+Cr2O3 (s) ->2Cr (s) +3CO and the enthalpy will be 1130-330 = 800 kJ/mol.