1/2 H2(g) + 1/2 I2(s) -> HI(g) ΔH = 26 kJ/molrxn 1/2 H2(g) + 1/2 I2(g) -> HI(g) ΔH = -5.0 kJ/molrxn — Thermodynamics Chemistry Question

STEPS:

1. Identify the target reaction and target species: The chemical equation representing the target process (the sublimation of iodine) is:
$$\text{I}_2(s) \rightarrow \text{I}_2(g)$$
We need to manipulate the two given chemical equations so that when added together, they yield this target equation. This application of enthalpy additivity is known as Hess's Law.
2. Manipulate the first given equation:
The first equation is:
$$\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{I}_2(s) \rightarrow \text{HI}(g) \quad \Delta H = 26\text{ kJ/mol}_{rxn}$$
Since our target reaction requires $1$ mole of solid iodine reactant ($\text{I}_2(s)$) on the left-hand side, we must multiply this entire equation and its corresponding enthalpy change by $2$:
$$\text{H}_2(g) + \text{I}_2(s) \rightarrow 2\text{HI}(g) \quad \Delta H = 2 \times (26\text{ kJ/mol}_{rxn}) = \mathbf{52\text{ kJ/mol}_{rxn}}$$
3. Manipulate the second given equation:
The second equation is:
$$\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{I}_2(g) \rightarrow \text{HI}(g) \quad \Delta H = -5.0\text{ kJ/mol}_{rxn}$$
Since our target reaction requires $1$ mole of gaseous iodine product ($\text{I}_2(g)$) on the right-hand side, we must reverse this equation (which changes the sign of its $\Delta H$ from negative to positive) and multiply the entire equation by $2$:
$$2\text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g) \quad \Delta H = -2 \times (-5.0\text{ kJ/mol}_{rxn}) = \mathbf{+10.0\text{ kJ/mol}_{rxn}}$$
4. Combine the manipulated equations and sum their enthalpies:
Add the two modified equations together:
$$\text{H}_2(g) + \text{I}_2(s) + 2\text{HI}(g) \rightarrow 2\text{HI}(g) + \text{H}_2(g) + \text{I}_2(g)$$
Cancel out the spectator species that appear on both sides of the reaction arrow ($\text{H}_2(g)$ and $2\text{HI}(g)$) to get our final target equation:
$$\text{I}_2(s) \rightarrow \text{I}_2(g)$$
According to Hess's Law, we sum the manipulated enthalpy values:
$$\Delta H = 52\text{ kJ/mol}_{rxn} + 10.0\text{ kJ/mol}_{rxn} = \mathbf{62\text{ kJ/mol}_{rxn}}$$
This matches Option E.

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WHY_OTHERS_WRONG:

• A is incorrect: A value of $15\text{ kJ/mol}_{rxn}$ is obtained if a student mistakenly subtracts the two values without reversing the second equation's sign or multiplying by the coefficient of 2 ($26 - 11$), representing a calculation and conceptual error.
• B is incorrect: A value of $21\text{ kJ/mol}_{rxn}$ is obtained if a student simply subtracts the two given enthalpy values ($26 - 5.0 = 21$) without accounting for the stoichiometric coefficient of $\frac{1}{2}$ or reversing the sign of the gaseous reactant.
• C is incorrect: A value of $31\text{ kJ/mol}_{rxn}$ represents the enthalpy change for the sublimation of only half a mole of iodine: $\frac{1}{2}\text{I}_2(s) \rightarrow \frac{1}{2}\text{I}_2(g)$ with $\Delta H = 26 - (-5.0) = 31\text{ kJ}$. The student forgot to multiply by 2 to get the enthalpy per full mole of $\text{I}_2$ as requested in the target equation.
• D is incorrect: A value of $42\text{ kJ/mol}_{rxn}$ represents a combination error where a student might multiply $26$ by 2 ($52$) but then subtract $10$ instead of adding it ($52 - 10 = 42$), representing a sign error where they failed to reverse the sign of the second reaction when turning it into a reverse step.