JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Application +4 | -1

In the reaction

\mathrm{CS}_2({l})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{SO}_2(\mathrm{g}) ; \Delta H= -265 \text{ kcal}.

The enthalpies of formation of

\mathrm{CO}_2

and

\mathrm{SO}_2

are both negative and are in the ratio

4: 3.

The enthalpy of formation of

\mathrm{CS}_2

+26 \mathrm{~kcal} \mathrm{~mol}^{-1}.

Calculate the enthalpy of formation of

\mathrm{SO}_2.

View answer and explanation

Correct option: D

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

Option D (

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

) is correct because it is calculated using Hess's law and the given enthalpy of reaction and formation of other compounds.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →

\ \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}