JEE Chemistry 2023 Question with Solution

Step-by-step solution

Standard Method

Given:

$$H_2(g) + O_2(g) \rightarrow 2OH(g), \quad \Delta H^\circ = 78 \, \text{kJ mol}^{-1}$$ $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g), \quad \Delta H^\circ = -242 \, \text{kJ mol}^{-1}$$ $$H_2(g) \rightarrow 2H(g), \quad \Delta H^\circ = 436 \, \text{kJ mol}^{-1}$$ $$\frac{1}{2}O_2(g) \rightarrow O(g), \quad \Delta H^\circ = 249 \, \text{kJ mol}^{-1}$$

Find: The value of $X$ for

$$H_2O(g) \rightarrow H(g) + OH(g)$$

Using Hess’s Law, derive the required reaction in steps.

First, break $H_2O(g)$ into atoms:

$$H_2O(g) \rightarrow H_2(g) + \frac{1}{2}O_2(g), \quad \Delta H^\circ = +242 \, \text{kJ mol}^{-1}$$ $$H_2(g) \rightarrow 2H(g), \quad \Delta H^\circ = 436 \, \text{kJ mol}^{-1}$$ $$\frac{1}{2}O_2(g) \rightarrow O(g), \quad \Delta H^\circ = 249 \, \text{kJ mol}^{-1}$$

Adding these gives:

$$H_2O(g) \rightarrow 2H(g) + O(g), \quad \Delta H^\circ = 242 + 436 + 249 = 927 \, \text{kJ mol}^{-1}$$

Now use formation of one $OH(g)$ from atoms. From

$$H_2(g) + O_2(g) \rightarrow 2OH(g), \quad \Delta H^\circ = 78 \, \text{kJ mol}^{-1}$$

and atomization values,

$$H_2(g) \rightarrow 2H(g), \quad \Delta H^\circ = 436 \, \text{kJ mol}^{-1}$$ $$O_2(g) \rightarrow 2O(g), \quad \Delta H^\circ = 2 \times 249 = 498 \, \text{kJ mol}^{-1}$$

So for

$$2H(g) + 2O(g) \rightarrow 2OH(g)$$ $$\Delta H^\circ = 78 - 436 - 498 = -856 \, \text{kJ mol}^{-1}$$

Hence for

$$H(g) + O(g) \rightarrow OH(g)$$ $$\Delta H^\circ = -428 \, \text{kJ mol}^{-1}$$

Combine this with

$$H_2O(g) \rightarrow 2H(g) + O(g), \quad \Delta H^\circ = 927 \, \text{kJ mol}^{-1}$$

to get

$$H_2O(g) \rightarrow H(g) + OH(g)$$ $$X = 927 - 428 = 499 \, \text{kJ mol}^{-1}$$

Therefore, the value of $X$ is $499$.

Using the extracted combined reaction

Given: The extracted solution also shows a combined route.

Reverse twice the water formation reaction:

$$2H_2O(g) \rightarrow 2H_2(g) + O_2(g), \quad \Delta H^\circ = +2 \times 242 = +484 \, \text{kJ mol}^{-1}$$

Use

$$H_2(g) + O_2(g) \rightarrow 2OH(g), \quad \Delta H^\circ = +78 \, \text{kJ mol}^{-1}$$

And use

$$H_2(g) \rightarrow 2H(g), \quad \Delta H^\circ = +436 \, \text{kJ mol}^{-1}$$

After cancellation, the net reaction becomes

$$2H_2O(g) \rightarrow 2H(g) + 2OH(g)$$

with

$$\Delta H^\circ = 484 + 78 + 436 = 998 \, \text{kJ mol}^{-1}$$

Therefore, for one mole,

$$H_2O(g) \rightarrow H(g) + OH(g)$$ $$X = \frac{998}{2} = 499 \, \text{kJ mol}^{-1}$$

So the correct numerical answer is $499$.

The answer key showing $12$ disagrees with the extracted solution working; the solution clearly supports $499$.

Common mistakes

• Using $436 + 249 = 685$ directly as the energy for $H_2O(g) \rightarrow 2H(g) + O(g)$ is incorrect because that omits reversing the formation of $H_2O(g)$ from $H_2(g)$ and $\frac{1}{2}O_2(g)$. You must first add $+242 \, \text{kJ mol}^{-1}$ to decompose water into its elements.

• Taking $78/2 = 39$ as the enthalpy for $H(g) + O(g) \rightarrow OH(g)$ is incorrect because $78 \, \text{kJ mol}^{-1}$ is for forming $OH$ from molecular reactants $H_2(g)$ and $O_2(g)$, not from isolated atoms. First account for atomization of $H_2$ and $O_2$, then derive the bond formation step.

• Forgetting to reverse the sign when reversing $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$ leads to a wrong result. When the reaction is reversed to decompose water, the enthalpy changes from $-242$ to $+242 \, \text{kJ mol}^{-1}$.