Question

Calculate ?H for the reaction C2H4(g) + 6 F2(g) ? 2 CF4(g) + 4 HF(g) given the following data:

Equation ?H (kJ)
a. H2(g) + F2(g) ? 2 HF(g) ?H = -537
b. C(s) + 2 F2(g) ? CF4(g) ?H = -680
c. 2 C(s) + 2 H2(g) ? C2H4(g) ?H = 52.3

Answer 1

To find the enthalpy change (ΔH) for the reaction, Hess's law was applied using the given reactions and their enthalpy changes. The calculations resulted in an ΔH of -3560.3 kJ, indicating the reaction is exothermic.

Step-by-step explanation:

To calculate the enthalpy change (ΔH) for the reaction C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g), you need to use Hess's law which states that you can add or subtract known enthalpy changes of reactions to find the ΔH for a reaction whose ΔH is unknown.

Using the given data:

• H2(g) + F2(g) → 2 HF(g), ΔH = -537 kJ
• C(s) + 2 F2(g) → CF4(g), ΔH = -680 kJ
• 2 C(s) + 2 H2(g) → C2H4(g), ΔH = 52.3 kJ

The desired reaction can be obtained with the following steps:

1. Reverse reaction c and multiply its ΔH by 1, since we need C2H4(g) on the reactant side.
2. Use reaction a twice, multiplying its ΔH by 4, since we need 4 HF on the product side.
3. Use reaction b twice, as we need 2 CF4 on the product side.

So, the ΔH for the desired reaction is:

ΔH = [2(-680 kJ) + 4(-537 kJ)] - [1(52.3 kJ)]

Calculating gives:

ΔH = [-1360 kJ + (-2148 kJ)] - [52.3 kJ]

ΔH = -3508 kJ - 52.3 kJ

ΔH = -3560.3 kJ

The reaction is exothermic as enthalpy change is negative, which means heat is released.