Chemical Thermodynamics: application problems for JEE+NEET

Solution — Step by Step n {gas} = (moles of gaseous products) - (moles of gaseous reactants) = 0 - 3 = -3. The water is liquid, so it doesn't count. Rearranging: Convert S to kJ/K: \Delta S = -0.326 kJ/K. Final answers: U \approx \mathbf{-564.6 kJ}, G \approx \mathbf{-474.9 kJ}. Why This Works H and \Delta U differ only because of the PV work done by gases. When gas moles decrease ($\Delta n Including liquid water in n {gas}. Only gaseous species count when applying H = \Delta U + \Delta n RT. If the products were 2H 2O(g), \Delta n would be 2 - 3 = -1, giving a different \Delta U. Forgetting unit consistency for S. JEE often gives \Delta S in J/K but \Delta H in kJ. Convert one before applying \Delta G = \Delta H - T\Delta S.

Chemical Thermodynamics: Application Problems (3)

Question

For the reaction $2H_2(g) + O_2(g) \to 2H_2O(l)$ , $\Delta H = -572$ kJ at $298$ K. Find $\Delta U$ . Then find $\Delta G$ if $\Delta S = -326$ J/K. Take $R = 8.314$ J/(mol·K).

Solution — Step by Step

$\Delta n_{\text{gas}} =$ (moles of gaseous products) $-$ (moles of gaseous reactants) $= 0 - 3 = -3$ .

The water is liquid, so it doesn’t count.

Rearranging:

\Delta U = \Delta H - \Delta n RT

\Delta n RT = (-3)(8.314)(298) = -7432.7 \text{ J} \approx -7.43 \text{ kJ}

\Delta U = -572 - (-7.43) = -564.57 \text{ kJ}

\Delta G = \Delta H - T\Delta S

Convert $\Delta S$ to kJ/K: $\Delta S = -0.326$ kJ/K.

\Delta G = -572 - 298 \times (-0.326) = -572 + 97.15 = -474.85 \text{ kJ}

Final answers: $\Delta U \approx \mathbf{-564.6 \text{ kJ}}$ , $\Delta G \approx \mathbf{-474.9 \text{ kJ}}$ .

Why This Works

$\Delta H$ and $\Delta U$ differ only because of the $PV$ work done by gases. When gas moles decrease ( $\Delta n < 0$ ), the surroundings do work on the system, so $\Delta U$ is less negative than $\Delta H$ — exactly what we got.

The negative $\Delta G$ confirms the reaction is spontaneous at $298$ K. Even though $\Delta S < 0$ (entropy decreases — fewer moles of gas, ordered liquid water), the strongly exothermic enthalpy term dominates.

Alternative Method

Compute $\Delta U$ via bond energies: form $4$ O-H bonds ( $\sim 463$ kJ/mol each), break $2$ H-H ( $\sim 436$ ) and $1$ O=O ( $\sim 498$ ). Approximate: $4(463) - [2(436) + 498] = 1852 - 1370 = 482$ kJ released. Close-ish to our answer; bond energies are approximate.

Common Mistake

Including liquid water in $\Delta n_{\text{gas}}$ . Only gaseous species count when applying $\Delta H = \Delta U + \Delta n RT$ . If the products were $2H_2O(g)$ , $\Delta n$ would be $2 - 3 = -1$ , giving a different $\Delta U$ .

Forgetting unit consistency for $\Delta S$ . JEE often gives $\Delta S$ in J/K but $\Delta H$ in kJ. Convert one before applying $\Delta G = \Delta H - T\Delta S$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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