JEE CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

The enthalpy changes for the following processes are listed below :

\begin{array}{ll}\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{Cl}(g), & 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(g) \rightarrow 2 \mathrm{I}(g), & 151.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{ICl}(g) \rightarrow \mathrm{I}(g)+\mathrm{Cl}(g), & 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(s) \rightarrow \mathrm{I}_2(g), & 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}

Given that the standard states for iodine and chlorine are

\mathrm{I}_2(s)

and

\mathrm{Cl}_2(g)

, the standard enthalpy of formation for

\operatorname{ICl}(g)

is :

View answer and explanation

Correct option: A

+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option A (+16.8 kJ mol⁻¹) is correct because it is calculated using Hess's Law and the given enthalpy changes.

More Questions from THERMODYNAMICS

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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}