JEE CHEMISTRY MCQ: THERMODYNAMICS

1 Hard

1 of 1

Analyzing +4 | -1

The data below refers to gas phase reaction at constant pressure at

25^{\circ} \mathrm{C}.

\begin{aligned} & \text{(a) } \mathrm{CH}_3-\mathrm{CH}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2+\mathrm{H}; \Delta H_1=+420 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \text{(b) } \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}; \Delta H_2=+168 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

From these data, the enthalpy change

\Delta H

for the reaction

2 \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3+\mathrm{CH}_2=\mathrm{CH}_2

View answer and explanation

Correct option: C

-252 \mathrm{~kJ}

Option C (

-252 \mathrm{~kJ}

) is correct because the enthalpy change is calculated by subtracting the enthalpy of reaction (a) from reaction (b).

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}