Question

using the equations
$ce{h2(g) + f2(g) -> 2 hf(g)}$ \t $delta h^circ = -79.2 mathrm{kj/mol}$
$ce{c(s) + 2 f2(g) -> cf4(g)}$ \t $delta h^circ = +141.3 mathrm{kj/mol}$
determine the molar enthalpy (in kj/mol) for the reaction
$ce{c(s) + 4 hf(g) -> cf4(g) + 2 h2(g)}$

Explanation:

Step1: Reverse the first equation and multiply by 2

The first equation is $\ce{H2(g) + F2(g) -> 2HF(g)}$ with $\Delta H^\circ = -79.2\ \text{kJ/mol}$. Reversing it gives $\ce{2HF(g) -> H2(g) + F2(g)}$ and the enthalpy change becomes $\Delta H_1^\circ = +79.2\ \text{kJ/mol}$. Multiplying this reversed equation by 2: $\ce{4HF(g) -> 2H2(g) + 2F2(g)}$, and the enthalpy change is $\Delta H_1^\circ = 2\times(+79.2) = +158.4\ \text{kJ/mol}$.

Step2: Use the second equation as is

The second equation is $\ce{C(s) + 2F2(g) -> CF4(g)}$ with $\Delta H_2^\circ = +141.3\ \text{kJ/mol}$.

Step3: Add the two modified equations

Adding the equation from Step1 ($\ce{4HF(g) -> 2H2(g) + 2F2(g)}$) and the equation from Step2 ($\ce{C(s) + 2F2(g) -> CF4(g)}$) gives:
$\ce{C(s) + 4HF(g) -> CF4(g) + 2H2(g)}$
Now, calculate the total enthalpy change: $\Delta H^\circ = \Delta H_1^\circ + \Delta H_2^\circ = +158.4 + 141.3 = +300.0 - 58.7$? Wait, no, wait. Wait, Step1: original first equation: $\ce{H2 + F2 -> 2HF}$, $\Delta H = -79.2$. So reverse: $\ce{2HF -> H2 + F2}$, $\Delta H = +79.2$. Multiply by 2: $\ce{4HF -> 2H2 + 2F2}$, $\Delta H = 2\times79.2 = 158.4$. Then second equation: $\ce{C + 2F2 -> CF4}$, $\Delta H = 141.3$. Now add the two: $\ce{4HF + C + 2F2 -> 2H2 + 2F2 + CF4}$. Cancel 2F2: $\ce{C + 4HF -> CF4 + 2H2}$. So total $\Delta H = 158.4 + 141.3$? Wait, no, wait: the first reversed equation (after multiplying by 2) has $\Delta H = +158.4$, and the second has $\Delta H = +141.3$? Wait, no, original first equation: $\ce{H2 + F2 -> 2HF}$, $\Delta H = -79.2$. So when we reverse it, $\ce{2HF -> H2 + F2}$, $\Delta H = +79.2$ (because reversing a reaction changes the sign of $\Delta H$). Then multiplying by 2: $\ce{4HF -> 2H2 + 2F2}$, so $\Delta H = 2\times(+79.2) = +158.4$. Then the second equation is $\ce{C + 2F2 -> CF4}$, $\Delta H = +141.3$. Now, when we add the two equations: $\ce{4HF -> 2H2 + 2F2}$ (ΔH=+158.4) and $\ce{C + 2F2 -> CF4}$ (ΔH=+141.3). Adding them: $\ce{4HF + C + 2F2 -> 2H2 + 2F2 + CF4}$. The 2F2 cancels, giving $\ce{C + 4HF -> CF4 + 2H2}$. Now, the total ΔH is the sum of the two ΔH values: 158.4 + 141.3? Wait, no, that can't be. Wait, no, the first equation's reversed and multiplied by 2: ΔH is +158.4, and the second equation's ΔH is +141.3. But when we add them, the F2 cancels. Wait, but let's check the target reaction: $\ce{C + 4HF -> CF4 + 2H2}$. So let's see:

First equation (original): $\ce{H2 + F2 -> 2HF}$, ΔH = -79.2. So to get 4HF on the left, we need to reverse and multiply by 2: $\ce{4HF -> 2H2 + 2F2}$, ΔH = +158.4 (since reversing flips the sign, multiplying by 2 multiplies the ΔH by 2).

Second equation: $\ce{C + 2F2 -> CF4}$, ΔH = +141.3.

Now, add the two equations:

$\ce{4HF -> 2H2 + 2F2}$ (ΔH=+158.4)

$\ce{C + 2F2 -> CF4}$ (ΔH=+141.3)

Adding them: $\ce{4HF + C + 2F2 -> 2H2 + 2F2 + CF4}$

Cancel 2F2: $\ce{C + 4HF -> CF4 + 2H2}$

Now, the total ΔH is 158.4 + 141.3? Wait, no, that would be 300, but that seems high. Wait, no, wait: original first equation: ΔH is -79.2, so reversing gives +79.2, multiplying by 2 gives +158.4. Second equation: +141.3. So total ΔH is 158.4 + 141.3 = 300? Wait, no, 158.4 + 141.3 = 299.7? Wait, 158.4 + 141.3: 158 + 141 = 299, 0.4 + 0.3 = 0.7, so 299.7. Wait, but let's check again.

Wait, maybe I made a mistake in the sign. Let's re-express the target reaction:

Target: $\ce{C(s) + 4HF(g) -> CF4(g) + 2H2(g)}$

Let's denote the first equation as Eq1: $\ce{H2(g) + F2(g) -> 2HF(g)}$, ΔH1 = -79.2

Second equation as Eq2: $\ce{C(s) + 2F2(g) -> CF4(g)}$, ΔH2 = +141.3

We need to manipulate Eq1 to get 4HF on the left. So reverse Eq1: $\ce{2HF(g) -> H2(g) + F2(g)}$, ΔH1' = +79.2 (reverse, so sign flips)

Multiply reversed Eq1 by 2: $\ce{4HF(g) -> 2H2(g) + 2F2(g)}$, ΔH1'' = 2*ΔH1' = 2*79.2 = 158.4

Now, Eq2 is $\ce{C(s) + 2F2(g) -> CF4(g)}$, ΔH2 = 141.3

Now, add Eq1'' and Eq2:

$\ce{4HF(g) -> 2H2(g) + 2F2(g)}$ (ΔH=158.4)

$\ce{C(s) + 2F2(g) -> CF4(g)}$ (ΔH=141.3)

Adding them: $\ce{4HF(g) + C(s) + 2F2(g) -> 2H2(g) + 2F2(g) + CF4(g)}$

Cancel 2F2(g) from both sides: $\ce{C(s) + 4HF(g) -> CF4(g) + 2H2(g)}$, which is the target reaction.

Now, the total ΔH is ΔH1'' + ΔH2 = 158.4 + 141.3 = 299.7? Wait, 158.4 + 141.3: 158.4 + 140 = 298.4, +1.3 = 299.7. Wait, but let's check with another approach.

Alternatively, use Hess's law. The target reaction can be obtained by Eq2 - 2*Eq1.

Because Eq2 is $\ce{C + 2F2 -> CF4}$, ΔH2 = 141.3

2*Eq1 is $\ce{2H2 + 2F2 -> 4HF}$, ΔH = 2*(-79.2) = -158.4

So Eq2 - 2*Eq1 is $\ce{C + 2F2 - 2H2 - 2F2 -> CF4 - 4HF}$, which simplifies to $\ce{C + 4HF -> CF4 + 2H2}$, which is the target reaction.

Then ΔH = ΔH2 - 2*ΔH1 = 141.3 - 2*(-79.2) = 141.3 + 158.4 = 299.7 kJ/mol.

Ah, that's a better way. So using Hess's law, subtract 2 times the first equation from the second equation. Since subtracting a reaction is equivalent to adding its reverse, so ΔH = ΔH2 + (-2*ΔH1) = 141.3 + 2*79.2 = 141.3 + 158.4 = 299.7 kJ/mol.

Answer:

299.7