Calculate the enthalpy of formation if 78.5 g of carbon dioxide in the following reaction:
C(s) + H2O(g) --> CO2(g)
Use the following equations:
a) H2O(l) --> H2(g) + (1/2)O2(g): Δ°f = +285.8 kJ/mol
b) C2H6(g) --> 2C(s) + 3H2(g): Δ°f = +84.7 kJ/mol
c) 2CO2(g) + 3O2(g) --> C2H6(g) + (7/2)O2(g): Δ°f = +1560.7kJ/mol
I don't believe you can do that.
C(s) + H2O(g) ==> CO2(g)
a. The equation is not balanced.
b. In those equations to be used you don't have H2O(g) anywhere but only H2O(l)
c. The equation you have posted isn't balanced. You're OK with C but O doesn't balance and there's no H on the left. Try again.
To calculate the enthalpy of formation of carbon dioxide, we need to break down the given reaction into a combination of the provided equations and rearrange them to cancel out the unwanted reactants and products. Here's how you can do it step by step:
1. Start with the given reaction: C(s) + H2O(g) → CO2(g)
2. Notice that equation (a) involves H2O, which is already present in the given reaction. We need to cancel out H2O and convert it into CO2. Multiply equation (a) by 2 to balance the oxygen atoms:
2H2O(l) → 2H2(g) + O2(g)
3. We also need to convert equation (a) to gaseous state by using Hess's Law (enthalpy change is independent of the physical state):
2H2O(g) → 2H2(g) + O2(g)
4. Equation (b) contains carbon (C) and hydrogen (H2) that we need to cancel out. Multiply equation (b) by 2 and reverse it:
2C2H6(g) → 4C(s) + 6H2(g)
5. Equation (c) allows us to convert CO2 to C2H6. Multiply equation (c) by 2 and reverse it:
2C2H6(g) + 7O2(g) → 4CO2(g) + (7/2)O2(g)
6. Now, we can add all the equations together:
2H2O(g) + 2C2H6(g) + 7O2(g) → 2H2(g) + O2(g) + 4C(s) + 4CO2(g) + (7/2)O2(g)
7. Simplify the equation by canceling out common reactants and products:
2H2O(g) + 2C2H6(g) + 7O2(g) → 2H2(g) + 4C(s) + 5CO2(g) + (7/2)O2(g)
8. Now, calculate the enthalpy change for this equation by adding the enthalpies of formation for the reactants and subtracting the enthalpies of formation for the products. The enthalpies of formation values are given:
ΔH° = (2 * Δ°f[CO2]) + (2 * Δ°f[C2H6]) - (2 * Δ°f[H2O]) - Δ°f[H2]
Substitute the values:
ΔH° = (2 * 1560.7 kJ/mol) + (2 * 84.7 kJ/mol) - (2 * 285.8 kJ/mol) - 0 kJ/mol
9. Calculate the total enthalpy change:
ΔH° = 3121.4 kJ/mol + 169.4 kJ/mol - 571.6 kJ/mol
ΔH° = 2719.2 kJ/mol
Hence, the enthalpy of formation for 78.5 g of carbon dioxide in the given reaction is 2719.2 kJ/mol.