Question

Consider the reaction at 300 K,

$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right);\Delta H=-185kJ/mol$.

Calculate $\Delta U$ if 3 mole of $H{}_2$ completely reacts with 3 mole of $C{l}_2$ to form $HCl$.

• A. $0$
• B. $-185$ kJ
• C. $-555$ kJ
• D. None of these

Answer 1

Answer: (C)

$-555$ kJ

Given:-

${H_2}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}⟶2{HCl}_{\left(g\right)};\Delta H=-185kJ$

Given that $3$ mole of $H_2$ reacts with $3$ moles of ${Cl}_2$.

Therefore

$3\times [{H_2}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}⟶2{HCl}_{\left(g\right)};\Delta H=-185kJ]$

$3{H_2}_{\left(g\right)}+3{{Cl}_2}_{\left(g\right)}⟶6{HCl}_{\left(g\right)};\Delta H=-555kJ$

For the above reaction,

$\Delta n_g=6-(3+3)=0$

A we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\because \Delta n_g=0$

$\therefore \Delta U=\Delta H=-555kJ$

Hence the value of $\Delta U=-555kJ$.