Consider the following reaction: 2 H_2(~g)+O_2(~g) arrow 2 H_2 O(g) Δ_ r H^ =-483.64 ~kJ . What is the enthalpy change for decomposition of one mole of water? (Choose the right option).

Correct answer: D. 241.82 ~kJ 2 H_2 O(g) 2 H_2(~g)+O_2(~g) Δ H^ =483.64 ~kJ For decomposition of 2 moles of H_2 O , enthalpy change =483.64 ~kJ For decomposition of 1 mole of H_2 O , enthalpy change =(483.64)/(2) ~kJ =241.82 ~kJ

NEET 2023

Exam: NEET
Year: 2023
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2023 · Chemistry · Thermodynamics (C)

Consider the following reaction: 2 H_2(~g)+O_2(~g) arrow 2 H_2 O(g)...

Consider the following reaction:

2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-483.64 \mathrm{~kJ}

. What is the enthalpy change for decomposition of one mole of water? (Choose the right option).

Options

A $18 \mathrm{~kJ}$
B $100 \mathrm{~kJ}$
C $120.9 \mathrm{~kJ}$
D $241.82 \mathrm{~kJ}$

Correct answer

D. $241.82 \mathrm{~kJ}$

Step-by-step solution

2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})

\Delta \mathrm{H}^{\circ}=483.64 \mathrm{~kJ}

For decomposition of 2 moles of

\mathrm{H}_2 \mathrm{O}

, enthalpy change

=483.64 \mathrm{~kJ}

For decomposition of 1 mole of

\mathrm{H}_2 \mathrm{O}

, enthalpy change

=\frac{483.64}{2} \mathrm{~kJ}

=241.82 \mathrm{~kJ}

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