Question

(1) MgCO3(s) ⟶ MgO(s) + CO2(g) ΔrH = 118.1 kJ mol−1

(2) Mg(s) + 12O2(g) ⟶ MgO(s) ΔfH =−601.7 kJ mol−1

What is the enthalpy change, ΔrH for the following reaction?
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)

Answer 1

I didn't write this down but tried to do it in my head; therefore, I may have made a mistake so check this out thoroughly.
Multiply eqn 1 by 2 and add it to the reverse of equn 2 multiplied by 2. See if that will give you the final equation you want. Change the sign of dH when reversing.

Answer 2

so, 118x-601 + 2x601?

Answer 3

no i dont think thats right

Answer 4

You're right, it isn't right. For two reasons.
First your second equation is not balanced.
Since it isn't balanced those instructions I gave can't be right.
Also you didn't follow instructions and check the final equation. That equation was not correct.
Multiply eqn 1 by 2 and add to the reverse of eqn 2.
Here is 1 and corrected 2.
(1) MgCO3(s) ⟶ MgO(s) + CO2(g) ΔrH = 118.1 kJ mol−1
(2) 2Mg(s) + O2(g) ⟶ 2MgO(s) ΔfH =−601.7 kJ mol−1 and you want
2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g)
Follow the instructions.
eqn1 x 2. 2MgCO3(s) ⟶ 2MgO(s) + 2CO2(g) ΔrH =118.1 kJ mol−1 x 2 =?
rev eqn 2. 2MgO(s) ==> 2Mg(s) + O2(g) −601.7 kJ mol−1 x -1 = ?
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add to get 2 MgCO3 ⟶ 2 Mg(s) + 2 CO2(g) + O2(g) (118.1*2) + 601,7 = ?