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  • For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, <img alt="\Delta_{\mathrm{C}} \mathrm{H}^{\ominus}=-601.70 \mathrm{~kJ} \mathrm{~mol}^{-1}" src="https

For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, \Delta_{\mathrm{C}} \mathrm{H}^{\ominus}=-601.70 \mathrm{~kJ} \mathrm{~mol}^{-1}, the magnitude of change in internal energy for the reaction is _____________kJ.(Nearest integer)

\text { (Given : } \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text { ) }

Option: 1

600


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Given,


        \begin{aligned} &\text{T}=300 \mathrm{~K} \\ \\&\text{P=1} \; \mathrm{bar} \\ \\&\mathrm{\Delta_{C} H^{0}=-601.7 \mathrm{~kJ} \mathrm{~mol}^{-1} }\\ \\&\mathrm{R=8.3 \mathrm{~J} \mathrm{~K}^{-1 }\mathrm{~mol}^{-1}} \end{aligned}

Reaction of Combustion-

\mathrm{M g(s)+\frac{1}{2} O_{2}(g) \longrightarrow M g O(s)} \\

\mathrm{\Delta H=\Delta U+\Delta n g R T} \\

\mathrm{-601.7 =\Delta U(kJ\; mol^{-1})-\frac{1\times 8.3 \times 300}{2\times 1000} } \\

On solving-

                \mathrm{\Delta U=-600.455 \mathrm{~kJ} \approx-600 \mathrm{~kJ}}

Hence, 600 is the answer

Posted by

avinash.dongre

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