Question

Consider the reaction at $300K$ $C_6{H}_6\left(\ell \right)+\frac{15}{2}{O}_2\left(g\right)⟶{6CO}_2\left(g\right)+{3H}_{2}O\left(\ell \right);\Delta H=-3271kJ$

What is$\Delta U$ for the combustion of $1.5$ mole of benzene at ${27}^{\circ }C$?

• A. $-3267.25kJ$
• B. $-4900.88kJ$
• C. $-4906.5kJ$
• D. $-327.754kJ$

Answer 1

The correct option is C $-4906.5kJ$

$C_6{H}_6+\frac{15}{2}{O_2}_{\left(g\right)}⟶{6C{O}_2}_{\left(g\right)}+3H_{2}O,\Delta H=-3271kJ$

For this reaction $\Delta U=\Delta H$ at constant pressure & volume.

$\Delta U=\Delta H$

For $1.5$ mole benzene

$\Delta U=1.5\times \Delta H$

$=1.5\times -3271kJ$

$=-4906.5kJ$.