Question

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction:
$2LiOH+C{O}_2\left(g\right)⟶{Li}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)$
$\Delta H_{f}LiOH\left(s\right)=-487.23kJ/mole$
$\Delta H_f{Li}_{2}C{O}_3\left(s\right)=-1215.6kJ/mole$
$\Delta H_f{H}_{2}O\left(l\right)=-285.85kJ/mole$
$\Delta H_{f}C{O}_2\left(g\right)=-393.5kJ/mole$

• A. $+303.4$
• B. $-133.5$
• C. $-198.6$
• D. $+198.6$

Answer 1

The correct option is B $-133.5$

$2LiOH+C{O}_2\left(g\right)⟶{Li}_{2}C{O}_3\left(s\right)+H_{2}O\left(l\right)$
$\Delta H_{f}LiOH\left(s\right)=-487.23kJ/mole$
$\Delta H_f{Li}_{2}C{O}_3\left(s\right)=-1215.6kJ/mole$
$\Delta H_f{H}_{2}O\left(l\right)=-285.85kJ/mole$
$\Delta H_{f}C{O}_2\left(g\right)=-393.5kJ/mole$

As we know,

Heat of reaction = Heat of formation of products - Heat of formation of reactants

$=(-1215.6-285.85)-\left(\right(2\times -487.23)+(-393.5\left)\right)=-133.5$ kJ/mole