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  • For the reaction <img alt="\begin{aligned} &\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \\ &\Delta U=-59.6 \mat

For the reaction

\begin{aligned} &\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \\ &\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { at } 27^{\circ} \mathrm{C} . \end{aligned}

The enthalpy change for the above reaction is \mathrm{(-)} __________ \mathrm{\mathrm{kJ} \mathrm{mol}^{-1}} [nearest integer]

\mathrm{\text { Given: } \mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text {. }}

Option: 1

57


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The Reaction

\mathrm{\mathrm{H}_2 \mathrm{~F}_2(\mathrm{g}) \longrightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(g)\; \; \; \; \quad \Delta H=?}

So, \mathrm{\begin{aligned} \mathrm{\Delta ng =2-1=1} \\ \end{aligned}}

       \mathrm{\Delta V=-59.6\; KJ\; mol^{-1}}

       \mathrm{T=27^{o}C=300\; K}

        \mathrm{R=8.314=J/mol-k}

We know, 

\mathrm{\Delta H=\Delta U+\Delta n g R T}                        \mathrm{\begin{aligned} &\mathrm{\Delta H=-50.6 \times 10^3+1 \times 8.314 \times 300} \\ &\mathrm{\Delta H=-57.105 \times 10^3 \mathrm{~J} / \mathrm{mol}} \\ &\mathrm{\Delta H=-57 \mathrm{~kJ} / \mathrm{mol}} \end{aligned}}

Answer = 57

Posted by

Deependra Verma

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