Question

Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction: $TiC{L}_4\left(g\right)+2H_{2}O\left(g\right)\to Ti{O}_2\left(g\right)+4HCl\left(g\right)$

$\Delta H_f^{\circ }TiC{L}_4\left(g\right)=-763.2kJ/mol$
$\Delta H_f^{\circ }Ti{O}_2\left(g\right)=-944.7kJ/mol$
$\Delta H_f^{\circ }{H}_{2}O\left(g\right)=-241.8kJ/mol$
$\Delta H_f^{\circ }HCl\left(g\right)=-92.3kJ/mol$

• A. $-278.1$
• B. $+369.2$
• C. $+67.1$
• D. $-67.1$

Answer 1

The correct option is D $-67.1$

The enthalpy change for the reaction is the difference in the enthalpies of formation of products and the enthalpies of formation of reactants.

$\Delta H_r=\left[\right(\Delta H_f{)}_{Ti{O}_2}+4(\Delta H_f{)}_{HCl}-[(\Delta H_f{)}_{TiC{l}_4}+2(\Delta H_f{)}_{H_{2}O}]$

Substitute values in the above expression.

$\Delta H_r=-944.7+(4\times -92.3)-[-763.2+(2\times -241.8)$

$\Delta H_r=-67.1kJ/mol$

Hence, the heat of reaction for the hydrolysis of titanium tetrachloride is - 67.1 kJ/mol.