JEE Main 2023 · 1 February · morning · Numerical answer · +4 / −0
At 25° C, the enthalpy of the following processes are given :
What would be the value of X for the following reaction ? _____________ (Nearest integer)
H₂O(g)→ H(g)+OH(g) Δ H°=X kJ mol⁻¹
| H₂(g)+O₂(g) | → | 2OH(g) | Δ H°=78 kJ mol⁻¹ |
|---|---|---|---|
| H₂(g)+1/2O₂(g) | → | H₂O(g) | Δ H°=-242 kJ mol⁻¹ |
| H₂(g) | → | 2H(g) | Δ H°=436 kJ mol⁻¹ |
| 1/2O₂(g) | → | O(g) | Δ H°=249 kJ mol⁻¹ |
Answer and solution
Answer 499
Solution
Chemistry ma’am · 3 steps
Build the target from the given steps: reverse water formation, then take half of the other two.
- Plan the route. First break H₂O(g) back into H₂ and O₂. Then make one OH and one H from those pieces. By Hess's law, Δ H for the target is the sum of Δ H for these steps.
Step used What we did Δ H° (kJ mol⁻¹) H₂O(g) → H₂(g) + 1/2O₂(g) reversed reaction 2, so the sign flips +242 1/2H₂(g) + 1/2O₂(g) → OH(g) half of reaction 1 78/2 = +39 1/2H₂(g) → H(g) half of reaction 3 436/2 = +218 - Check the sum. Add the three equations. The H₂ and 1/2O₂ made in step 1 are used up in steps 2 and 3, leaving H₂O(g) → H(g) + OH(g). The O(g) data is not needed.
X = 242 + 39 + 218 = 499
- So. Δ H° = 499 kJ mol⁻¹, so X = 499.
The common slip: 756. Adding the full 78 and 436 without halving them gives 242 + 78 + 436 = 756. The target needs only one OH and one H.
The idea to keep. In Hess's law problems, write the target first, then flip and scale the given equations until the extra species cancel. Flipping changes the sign, and scaling scales Δ H too.
Answer from the NTA final answer key. Solution by PrepWiser.