${{\rm{H}}_2}\left( {\rm{g}} \right) + \frac{1}{2}{{\rm{O}}_2}\left( {\rm{g}} \right) \to {{\rm{H}}_2}{\rm{O}}\left( l \right);\Delta H = - 68{\rm{kcal}}$.The heat change, for the decomposition of 3.6 g of water is :

JEE CHEMISTRY MCQ: THERMODYNAMICS

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{\mathrm{H}_2}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right) \to {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 68\mathrm{kcal}

.The heat change, for the decomposition of 3.6 g of water is :

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Correct option: B

13.6 kcal

Option B (13.6 kcal) is correct because the heat change is calculated based on the stoichiometry of the reaction and the given enthalpy change.

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\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

0.18 \mathrm{~g}

4 \mathrm{~K}.

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

{CsI}

{CsCl}

{CsCl}

{Cs} I

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

\Delta H

\Delta u

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}