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  • For complete combustion of methanol <img alt="\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathr

For complete combustion of methanol
\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})
the amount of heat produced as measured by bomb calorimeter is726 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}. The enthalpy of combustion for the reaction is -x \mathrm{~kJ} \mathrm{~mol}^{-1}, where x is _____________.(Nearest integer)
(Given : \mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} )

Option: 1

727


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\mathrm{CH}_{3} \mathrm{OH}{(\ell)}+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)

\mathrm{Given, \Delta U=-726 \mathrm{KJ} /\mathrm{mol}
                \mathrm{T =27^{\circ} \mathrm{C}=300 \mathrm{~K}}
                 \mathrm{R =8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}

\mathrm{\Delta n_{g} =1-\frac{3}{2}=-\frac{1}{2}} 

Now, the relation between \mathrm{ \Delta U} and \mathrm{ \Delta H} is given as 

\mathrm{\Delta H =\Delta U+(\Delta n_ g) \, R T}

\Rightarrow\mathrm{\Delta H=\left(-726-\frac{1}{2} \times \frac{8.3 \times 300}{1000}\right) \mathrm{kJmol}^{-1}}
\Rightarrow \mathrm{\Delta H}\mathrm{=-727.2 \mathrm{~kJ} \mathrm{~mol}^{-1}}

\mathrm{Value\: of\: x\: is\: \; 727 .}

Hence, the answer is 727

Posted by

Ramraj Saini

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