Question

Given that, the molar combustion enthalpy of benzene, cyclohexane and hydrogen are $x,y,$ and $z,$ respectively, the molar enthalpy of hydrogenation of benzene to cyclohexane is

• A. $x-y+z$
• B. $x-y+3z$
• C. $y-x+z$
• D. $y-x+3z$

Answer 1

Answer: (B)

Hydrogenation of benzene is given by
$C_6{H}_6+3H_2\to C_6{H}_{12}\dots \left(a\right)$
Enthalpies are given for,
$C_6{H}_6+\frac{15}{2}{O}_2\to 6C{O}_2+3H_{2}O;\Delta H=x\dots \left(b\right)$
$C_6{H}_{12}+9O_2\to 6C{O}_2+6H_{2}O;\Delta H=y\dots \left(c\right)$
$H_2+\frac{1}{2}{O}_2\to H_2;\Delta H=z\dots \left(d\right)$
Reaction$\left(a\right)=\left(b\right)-\left(c\right)+3\left(d\right),$
$\therefore \Delta H$ of $a=x-y+3z$