Question

Given the following bond energies,
$C-H=414KJ/mol$
$C-Cl=150KJ/mol$
$Cl-Cl=243KJ/mol$

$H-Cl=432KJ/mol$
How much energy would be required in the reaction?
${CH}_{4\left(g\right)}+{2Cl}_{2\left(g\right)}\to {CH}_2{Cl}_{2\left(g\right)}+{2HCl}_{\left(g\right)}$

• A. $-150KJ$
• B. $-130KJ$
• C. $-228KJ$
• D. $-571KJ$

Answer 1

The correct option is A $-150KJ$

$C{H}_4=4C-H$ bonds=$4\left(414\right)=1656kJ/mol$

$C{l}_2=1Cl-Cl$ bond= $243kJ/mol$

$C{H}_{2}C{l}_2=2C-H$ & $2C-Cl$ bonds

$=2\left(414\right)+2\left(150\right)=828+300=1128kJ/mol$

$HCl=1H-Cl$ bond= $432kJ/mol$

Energy required in the following reaction is

${C{H}_4}_{\left(g\right)}+{2C{l}_2}_{\left(g\right)}\to {C{H}_{2}C{l}_2}_{\left(g\right)}+2HC{l}_{\left(g\right)}$

$=(1128+2\times 432)-(2\times 243+1656)$

$=1128+864-(486+1656)$

$=1128+864-2142$

$=-150kJ$