The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction: AgNO3(aq)+HCl(aq)→AgCl(s)+HNO3(aq). When you combine 70.0mL of 0.185M AgNO3 with 70.0mL of 0.185M HCl in a coffee-cup calorimeter, the temperature changes from 23.16∘C to 24.25∘C.. Calculate ΔHrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and C=4.18J/g⋅∘C as the specific heat capacity.

C = 4.18 delta T = 24.25 - 23.16 = 1.09 mol = 0.070L * 0.185M = 0.01295 m = d*v = 1.00*(0.070+0.070) = 0.14g So, delta H = q/mol HCl = [mC(deltaT)]/mol HCl = (0.14*4.18*1.09)/(0.01295) = 49.3 kJ

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Аноним Аноним · Answer 2 · 2019-11-15T10:32:39+01:00

Answer: The enthalpy change of the reaction is 49305.02 J/mol

Explanation:

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

We are given:

Density of solution = 1.00 g/mL

Volume of solution = (70.0 + 70.0) mL = 140.0 mL

Putting values in above equation, we get:

[tex]1g/mL=\frac{\text{Mass of solution}}{140.0mL}\\\\\text{Mass of solution}=(1g/mL\times 140.0mL)=140.0g[/tex]

To calculate the heat absorbed, we use the equation:

[tex]q=mc\Delta T[/tex]

where,

q = heat absorbed

m = mass of solution = 140.0 g

c = specific heat capacity of water = 4.184 J/g.°C

[tex]\Delta T[/tex] = change in temperature = [tex][24.25-23.16]^oC=1.09^oC[/tex]

Putting values in above equation, we get:

[tex]q=140.0g\times 4.184J/g.^oC\times 1.09^oC\\\\q=638.5J[/tex]

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

Molarity of HCl solution = 0.185 M

Volume of solution = 70.0 mL

Putting values in above equation, we get:

[tex]0.185=\frac{\text{Moles of HCl}\times 1000}{70.0}\\\\\text{Moles of HCl}=\frac{0.185\times 70.0}{1000}=0.01295mol[/tex]

To calculate the enthalpy change of the reaction, we use the equation:

[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]

where,

[tex]q[/tex] = amount of heat absorbed = 638.5 J

n = number of moles = 0.01295 moles

[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=\frac{638.5J}{0.01295mol}=49305.02J/mol[/tex]

Hence, the enthalpy change of the reaction is 49305.02 J/mol