Question

Find the forward${\Delta }_r{U}^{\circ }$ at 300K for the reaction $4HCl\left(g\right)+O_2\left(g\right)\to 2C{l}_2\left(g\right)+2H_{2}O\left(g\right)$ Assume all gases are ideal. Given $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$ ${\Delta }_r{H}_{300}^{{}^{\circ }}=-184.5kJ/mol$
$2H_2\left(g\right)+O_2\left(g\right)\to 2H_{2}O\left(g\right)$ ${\Delta }_r{H}_{300}^{{}^{\circ }}=-483kJ/mol$ (Use R= 8.3 J/mole)

• A. 111.5 kJ/mole
• B. -109.01 kJ/mole
• C. -111.5 kJ/mole
• D. None

Answer 1

The correct option is C -111.5 kJ/mole

$4HCl\left(g\right)+O_2\left(g\right)⟶2C{l}_2\left(g\right)+2H_{2}O\left(g\right)......\left(i\right)$
$H_2\left(g\right)+C{l}_2\left(g\right)⟶2HCl\left(g\right)....\left(ii\right)$
$2H_2\left(g\right)+O_2\left(g\right)⟶2H_{2}O\left(g\right)....\left(iii\right)$
As it can be clearly seen
$Eq\left(i\right)=Eq\left(iii\right)-2Eq\left(ii\right)$
Thus $\Delta H_r^o\left(i\right)=\Delta H_r^o\left(iii\right)-2\Delta H_r^o\left(ii\right)$
$=-483-(2\times -184)$
$=-114kJ/mol$
$\Delta U_r^o=\Delta H_r^o-\Delta n_{g}RT$
Thus $\Delta U_r^o=-114-(4-5)\times 0.0083\times 300$
$\Rightarrow \Delta U_r^o=-111.5kJ/mol$