A strip of magnesium metal having a mass of 1.30 g dissolves in 150. ml of 5.00 M HCl (specific gravity = 1.10); products of the reaction are magnesium chloride and hydrogen gas. The HCl is initially at 22.0°C, and the resulting solution reaches a final temperature of 52.0°C. The heat capacity of the calorimeter in which the reaction occurs is 345 J/°C. Calculate ΔH (in kJ/mol) for the reaction under the conditions of the experiment, assuming the specific heat of the final solution is the same as that for water (4.184 J/g°C).

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OlaMacgregor OlaMacgregor · Answer 1 · 2019-10-07T07:35:54+02:00

Explanation:

The given data is as follows.

Mass of Mg = 1.30 g, Molarity = 5 for HCl, Volume = 150 mL

Density = 1.1 g/mL

Change in temperature (dT) = [tex](52 - 22)^{o}C = 30^{o}C[/tex]

[tex]C_{p}[/tex] = 345 [tex]J/^{o}C[/tex]

Hence, the energy balance will be as follows.

[tex]-Q_{rxn} = Q_{calorimeter} + Q_{water}[/tex]

[tex]Q_{rxn} = H_{rxn}/mol[/tex]

[tex]H_{Rxn} = Q_{Rxn}/mol[/tex]

Moles of Mg = [tex]\frac{mass}{\text{molecular mass}}[/tex]

= [tex]{1.3}{24.3}[/tex]

= 0.05349 mol of Mg

[tex]Q_{calorimter} = C \times dT[/tex]

= [tex]345 J/^{o}C \times 30^{o}C[/tex]

= 10350 J

[tex]Q_{water} = m \times C \times (T_{f} - T_{i})[/tex]

= [tex]150 \times 1.1 \times 4.184 \times 30^{o}C[/tex]

= 20710.8

[tex]-Q_{rxn}[/tex] = (20710.8 + 10350) J

[tex]Q_{rxn}[/tex] = -31060.8 J

Therefore, calculate the value of [tex]\Delta H[/tex] of the reaction as follows.

[tex]Q_{rxn} = H_{rxn}/mol[/tex]

[tex]H_rxn = \frac{-31060.8}{0.05349}[/tex]

[tex]H_{Rxn}[/tex] = -580684.24 J/mol

or, [tex]H_{Rxn} = -580.68 kJ/mol[/tex] (as 1 kJ = 1000 J)

Thus, we can conclude that value of [tex]\Delta H[/tex] of the reaction under given conditions is -580.68 kJ/mol.