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Enthalpy Change Formula: Find ΔH per Mole of Zinc

Zinc is limiting because 0.0450 mol HCl exceeds the 0.0410 mol required. Dividing the released heat, −3.14 kJ, by 0.0205 mol Zn gives ΔH = −153 kJ mol⁻¹ Zn.

Question

A 1.34 g1.34\,\mathrm g sample of Zn(s)\mathrm{Zn(s)} reacts with 60.0 mL60.0\,\mathrm{mL} of 0.750 M HCl(aq)0.750\,\mathrm M\ HCl(aq) according to

Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)}

and releases 3.14 kJ3.14\,\mathrm{kJ}. Determine ΔH\Delta H per mole of Zn consumed.

Answer

The enthalpy change is −153 kJ mol−1 Zn-153\,\mathrm{kJ\,mol^{-1}\ Zn}.

Convert zinc to moles:

n(Zn)=1.34 g65.38 g mol−1=0.0205 mol.n(\mathrm{Zn})=\frac{1.34\,\mathrm g}{65.38\,\mathrm{g\,mol^{-1}}} =0.0205\,\mathrm{mol}.

Check that HCl is sufficient:

n(HCl)=(0.0600 L)(0.750 mol L−1)=0.0450 mol.n(\mathrm{HCl})=(0.0600\,\mathrm L)(0.750\,\mathrm{mol\,L^{-1}}) =0.0450\,\mathrm{mol}.

The reaction needs 2(0.0205)=0.0410 mol HCl2(0.0205)=0.0410\,\mathrm{mol\ HCl}, so HCl is in excess and Zn is the limiting reactant.

Because the reaction releases heat, the system's heat is negative: qp=−3.14 kJq_p=-3.14\,\mathrm{kJ}. At constant pressure, ΔH=qp\Delta H=q_p. Normalize by the moles of zinc that react:

ΔHn(Zn)=−3.14 kJ0.0205 mol Zn=−153 kJ mol−1 Zn.\frac{\Delta H}{n(\mathrm{Zn})} =\frac{-3.14\,\mathrm{kJ}}{0.0205\,\mathrm{mol\ Zn}} =\mathbf{-153\,\mathrm{kJ\,mol^{-1}\ Zn}}.

Evidence boundary

This result assumes constant-pressure conditions and assigns the measured 3.14 kJ entirely to the stated reaction. It verifies Zn as the limiting reactant and reports the value per mole of Zn, as requested. The negative sign denotes heat released; no calorimeter or environmental heat-loss correction is available from the prompt.

Sources

These references support the concepts and methods used in the explanation above.

Enthalpy Change Formula per Mole of Zinc | Verla