Consider the following reaction: 2H2(g) + O2(g) -> 2H2O(g), _r H^° = -483.64 kJ. What is the enthalpy change for the decomposition of one mole of water?

Correct answer: (D) 241.82 kJ. Answer: (D) 241.82 kJ Solution: The formation of 2 mol of H2O(g) releases 483.64 kJ, so the formation of 1 mol releases 483.642 = 241.82 kJ ( H = -241.82 kJ,mol-1). Decomposition of 1 mol of water is the reverse process, so its enthalpy change is +241.82 kJ. Mapped: source numeric answer '4' → option D.

NEET 2023 Phase 1 · ChemistryPrevious Year Question

Consider the following reaction: $2 H X_{2} (g) + O X_{2} (g) 2 H X_{2} O (g)$ , $Δ_{r} H^{\circ} = - 483.64 kJ$ . What is the enthalpy change for the decomposition of one mole of water?

Answer: (D) $241.82 kJ$ . \textbf{Answer:} (D) $241.82 kJ$ \textbf{Solution:} The formation of $2$ mol of $H X_{2} O (g)$ releases $483.64 kJ$ , so the formation of $1$ mol releases $\frac{483.64}{2} = 241.82 kJ$ ( $Δ H = - 241.82 k J m o l^{- 1}$ ).

A. $18 kJ$
B. $100 kJ$
C. $120.9 kJ$
D. $241.82 kJ$ ✓

Correct Answer

(D) $241.82 kJ$

Solution & Explanation

\textbf{Answer:} (D) $241.82 kJ$ \textbf{Solution:} The formation of $2$ mol of $H X_{2} O (g)$ releases $483.64 kJ$ , so the formation of $1$ mol releases $\frac{483.64}{2} = 241.82 kJ$ ( $Δ H = - 241.82 k J m o l^{- 1}$ ). Decomposition of $1$ mol of water is the reverse process, so its enthalpy change is $+ 241.82 kJ$ . Mapped: source numeric answer '4' → option D.

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Consider the following reaction: 2HX2​(g)+OX2​(g)​2HX2​O(g), Δr​H∘=−483.64 kJ. What is the enthalpy change for the decomposition of one mole of water?

Solution & Explanation

More NEET PYQ solutions

Consider the following reaction: $2 H X_{2} (g) + O X_{2} (g) 2 H X_{2} O (g)$ , $Δ_{r} H^{\circ} = - 483.64 kJ$ . What is the enthalpy change for the decomposition of one mole of water?