Using the data below, calculate the enthalpy for the combustion of C to CO C(s) + O2(g)--> CO2 (g) ΔH1 = -393.5 kJ

CO(g) + 1/2O2(g) -->CO2(g) ΔH2= -283.0kJ

Target: C(s) + 1/2O2(g) --> CO(g) ΔH3=??

Group of answer choices

ΔH3 = -110.5 kJ.

Explanation:

Hello!

In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:

C(s) + 1/2O2(g) --> CO(g)

Thus, by letting the first reaction to be unchanged:

C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ

And the second one inverted:

CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ

If we add them, we obtain:

C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)

Whereas CO2 can be cancelled out and O2 subtracted:

C(s) + 1/2O2(g)  --> CO(g)

Therefore, the required enthalpy of reaction is:

ΔH3 = -393.5 kJ + 283.0kJ

ΔH3 = -110.5 kJ

Best regards!

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

56 moles of   lieter in chemistry