Question

128 Thermodynamics
21. Enthalpy changes for two reactions are given by
the equation
2Cr(s) + O2(g)
+ Cr2O3(s), AH = -1130 kJ
C(s) + ,(g)
+ CO(g),
AH = -110 kg.
The enthalpy change for the reaction
Cr20, + 3C 2Cr + 3CO will be
(1) -1460 kJ
(2) -800 kJ
(3) +800 kJ
(4) 1020 kJ​

Answer 1

Answer:

I think so

c)+800kj

Explanation:

_110×3=_330

and _11 30 will become + because reaction is reverse

Answer 2

Answer:

+800 kJ.

Explanation:

From the question we have that the first reaction is 2Cr (s) +3/2 O2 -> Cr2O3 , which is having a change in the enthalpy as ​ ∆H -1130kJ.

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

So, we 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.