Question

What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions?

Here's the reactions:

There was a formatting issue with the specific chemistry symbols, there all correct in the picture below

2C(s) + O2(g) → 2CO(g)


C(s) + O2(g) → CO2(g) ∆H0= -394 KJ/mol



CO(s) + 1/2 O2(g) → CO2(g) ∆H0= -283 KJ/mol


There was a formatting issue with the specific chemistry symbols, there all correct in the picture below

Answer 1

∆H0 = -222kJ/mol

Step-by-step explanation:

Using Hess's law, we can find the ΔH of a reaction from the sum of another related reactions as follows:

Using the reactions:

(1) C(s) + O2(g) → CO2(g) ∆H0= -394 KJ/mol

(2) CO(s) + 1/2 O2(g) → CO2(g) ∆H0= -283 KJ/mol

Twice (1):

2C(s) + 2O2(g) → 2CO2(g) ∆H0= 2*-394 KJ/mol = -788kJ/mol

The inverse reaction of (2):

-(2) CO2(g) → CO(g) + 1/2 O2(g) ∆H0= 283 KJ/mol

Twice this reaction:

2*-(2) 2CO2(g) → 2CO(s) + O2(g) ∆H0= 2*283 KJ/mol= 566kJ/mol

Now, the sum of 2*(1) - 2*(2) produce:

2C(s) + 2O2(g) + 2CO2(g)→ 2CO2(g) + 2CO(g) + O2(g) ∆H0= -788kJ/mol + 566kJ/mol

Subtracting the molecules that ar in both sides of the reaction:

2C(s) + O2(g) → 2CO(g) ∆H0 = -222kJ/mol