Using the following equation

Ca(s) + 2C(s) --> CaC2(s)
^ H = -62.8 kJ

CO2(g) --> C(s) + O2(g)
^ H = 393.5 kJ

CaCO3(s) + CO2 (g) --> CaC2(s) + 2 1/2O2(g)
^ H = 1538 kJ

Determine the heat of the reaction (in kJ) for: Ca(s) + C(s) + 3/2O2(g) --> CaCO3(s)

Consider these three reactions happening together:

CaC2(s)+ 2 1/2O2(g)-> CaCO3(s)+ CO2 (g)
CO2 (g) -> C(s) + O2(g)
Ca(s) +2C(s) -> CaC2(s)

The net reaction is
Ca(s) + C(s) + (3/2)O2 (g) -> CaCO3 (s)

Add or subtract the three heats of formation, ^H, as appropriate. Some are the reverse of the reactions you wrote. Reverse the sign of ^H in those cases.

You should end up the the heat of reaction for the final "net" reaction

I am surprised to see a question like this in ninth grade. I was taught it in graduate school!

To determine the heat of the reaction for the given equation, we can use Hess's Law, which states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps of the reaction.

First, let's write out the given equations along with their respective enthalpy values:

1) Ca(s) + 2C(s) --> CaC2(s) ΔH = -62.8 kJ
2) CO2(g) --> C(s) + O2(g) ΔH = 393.5 kJ
3) CaCO3(s) + CO2(g) --> CaC2(s) + 2 1/2O2(g) ΔH = 1538 kJ

Now, we need to manipulate these equations in order to cancel out the substances that appear on both sides of the desired equation and create the overall reaction:

1) Ca(s) + 2C(s) --> CaC2(s)
2) CO2(g) --> C(s) + O2(g)
3) CaCO3(s) + CO2(g) --> CaC2(s) + 2 1/2O2(g)

We can see that equation (1) involves the formation of CaC2(s) with a certain enthalpy change. We can also see that equation (3) involves the formation of CaC2(s) with a different enthalpy change. By manipulating equation (2), we can combine equations (1) and (3) to form the desired equation.

Multiply equation (2) by 2 to balance the number of carbon atoms:

4) 2CO2(g) --> 2C(s) + 2O2(g) ΔH = 2 * 393.5 kJ = 787 kJ

Multiply equation (4) by 2/3 to balance the number of oxygen atoms:

5) 4/3CO2(g) --> 4/3C(s) + 4/3O2(g) ΔH = 4/3 * 787 kJ ≈ 1049.33 kJ

Now, we can combine equations (1) and (5) to obtain the desired equation:

6) Ca(s) + C(s) + 3/2O2(g) --> CaCO3(s) ΔH = -62.8 kJ + 1049.33 kJ = 986.53 kJ

Therefore, the heat of the reaction for the given equation is approximately 986.53 kJ.