Question

Titanium tetrachloride\(Ti{\rm{C}}{{\rm{l}}_4}\) has a melting point of\({\rm{23}}{\rm{.2}}^\circ {\rm{C}}\) and has a ΔH fusion =\(9.37\)kJ/mol.

(a) How much energy is required to melt \(263.1\)g\(Ti{\rm{C}}{{\rm{l}}_4}\)?

(b) For\(Ti{\rm{C}}{{\rm{l}}_4}\) which will likely have the larger magnitude: ΔH fusion or ΔH vaporization? Explain your reasoning.

Short answer

(a) \({\rm{TiC}}{{\rm{l}}_{\rm{4}}} - 13\,kJ\)_{is the amount of heat needed to vaporize.}

(b)Heat of vaporization will have a larger magnitude

Step-by-step solution

To find the energy

(a) To convert from mass to moles, we need to divide its mass by molecular weight therefore, moles are:-

\({\rm{263}}{\rm{.1\;g}}\,{\rm{TiC}}{{\rm{l}}_{\rm{4}}}{\rm{ \times }}\frac{{{\rm{1\;mol}}}}{{{\rm{189}}{\rm{.9\;g}}}}{\rm{ = 1}}{\rm{.385\;mol}}\,{\rm{TiC}}{{\rm{l}}_{\rm{4}}}\)

Heat needed to vaporize this amount of \({\rm{TiC}}{{\rm{l}}_{\rm{4}}}\)

\(\begin{aligned}{}{\rm{n\Delta }}{{\rm{H}}_{{\rm{fusion }}}}{\rm{ = (1}}{\rm{.385\;mol)(9}}{\rm{.37\;kJ/mol)}}\\{\rm{ = 13\;kJ}}\end{aligned}\)

To explain the reasoning.

(b) It is most likely that heat of vaporization will have a larger magnitude mainly because of the fact that in the case of vaporization, the intermolecular interactions have to be completely overcome, whereas, in the case of melting, intermolecular forces need to be weakened or destroys only some of them.

Hence Heat needed to vaporize this amount of \({\rm{TiC}}{{\rm{l}}_{\rm{4}}} - 13kJ\)