Question

Calculate the enthalpy of reaction for the following reaction using Hess's Law and the equations provided below.

Target Reaction: Sr(s)+C(graphite)+1 1/2 O₂(g) -----> SrCO₃(s) H(degrees)=?

C(Graphite)+ O₂(g)-----> CO₂(g) H(degrees)= -349kJ

Sr(s)+ 1/2 O₂(G)-------> SrO(s) H(degrees)= -592kJ

SrO(s)+CO₂(g)-------> SrCo₃(s) H(degrees)= -234kJ

Answer 1

first of all, you have witten the target equation wrongly it should be

Sr (s) + (3/2) O2 (g) + C (graphite) ---> SrCO 3 (s)

(A) Sr (s) + (1/2) O2 (g) ---> SrO (s) dH = -592 kJ
(B) SrO (s) + CO2 (g) ---> SrCO 3 (s) dH = -234 kJ
(C) C (graphite) + O2 (g) ---> CO 2 (g) dH = -394 kJ

we use 1 equation (B) to provide the product , SrCO 3 (s)
we use 1 equation (A) to provide the Sr as a starting material
we use 1 equation (C) to provide the C as a starting material
(we are not concerned with providing the necessary O2, since it is provided from more than 1 source

combining 1 (A) & 1(B) & 1 (C) gives:
Sr + (1/2) O2 & SrO (s) + CO2 (g) & C (graphite) + O2 (g) ---> SrO (s) & SrCO 3 (s) & CO 2
we notice that
SrO (s) --> cancels out .--> SrO (s)
CO2 (g) --> cancels out .--> CO2 (g)

leavng only your desired equation
Sr (s) + (3/2) O2 (g) + C (graphite)--> SrCO 3 (s)


Hess's law says that if adding 1 (A) & 1(B) & 1 (C) gives you the equation you wish,
the combining the dH's from 1 (A) & 1(B) & 1 (C) gives you the dH you wish

-592 kJ & -234 kJ & -394 kJ = -1220 kJ
your answer is
dH = -1220 kJ