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You want to determine ΔH° for the reaction Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)....

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spock2 | Student, Undergraduate | Honors

Posted August 28, 2012 at 3:34 PM via web

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You want to determine ΔH° for the reaction Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g). To do so, you first determine the heat capacity...

...of a calorimeter using the following reaction, whose ΔH is known: NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(I)

ΔH° = -57.32KJ

Calculate the heat capacity of the calorimeter from these data:

Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH

Initial T of both solutions: 16.9°C

Maximun T recorded during reaction: 30.4°C

Density of resulting NaCl solution: 1.04 g/mL

c of 1.00 M NaCl(aq) = 3.93 J/g*K

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted October 5, 2013 at 7:47 AM (Answer #1)

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The volume of the final solution is V =50+50 =100 mL with a density of 1.04 g/mL which means a total mass of solution

 `m =rho*V =1.04*100 =104 g`

The balanced chemical equation of transformation is

`NaOH +HCl=> NaCl + H_2O`

Number of moles of NaOH entering is n(NaOH) = 2*50/1000 =0.1

Number of moles of HCl entering is n(HCl) =2*50/1000 =0.1 moles

For 0.1 mole of NaOH and 0.1 mole of HCl it results 0.1 mole of NaCl

The molecular mass of NaCl is M(NaCl) =23+35.5 =58.5 g

The molarity of the final solution is n =0.1 mole/V =0.1/0.1 = 1 mole/L

The heat released by the reaction (the enthalpy given in problem is in KJ/mole)

`Q =n*Delta(H) =0.1*57.32 =5.732 KJ`

The heat balance is

`m_(sol)*c_(sol)*Delta(T) +C_(cal)*Delta(T) =Q`

`C_(cal) = (5732-104*3.93*(30.4-16.9))/(30.4-16.9) = 15.87 J/K`

The heat capacity of the calorimeter is C =15.87 J/K

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