Question

Zinc metal reacts with hydrochloric acid according to this balanced equation. Zn(s) 2HCl(aq)→ZnCl2(aq) H2(g) When 0.114 g of Zn(s) is combined with enough HCl to make 52.4 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.0 ∘C to 23.9 ∘C.

Answer 1

Answer: The enthalpy change of the reaction is 239.2 kJ/mol

Step-by-step explanation:

• To calculate the mass of solution, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of solution = 1.0 g/mL

Volume of solution = 52.4 mL

Putting values in above equation, we get:

`1.0g/mL=\frac{\text{Mass of solution}}{52.4mL}\\\\\text{Mass of solution}=(1.0g/mL* 52.4mL)=52.4g`

• To calculate the heat absorbed we use the equation:

`q=mc\Delta T`

where,

q = heat absorbed = ?

m = mass of solution = 52.4 g

c = specific heat capacity of solution = 4.18 J/g.°C

`\Delta T` = change in temperature =
`(23.9-22.0)^oC=1.9^oC`

Putting values in above equation, we get:

`q=52.4g* 4.18J/g.^oC* 1.9^oC\\\\q=416.2J`

Heat absorbed by the solution = heat released by the reaction

• To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Given mass of zinc = 0.114 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

`\text{Moles of zinc}=(0.114g)/(65.4g/mol)=0.00174mol`

• Calculating the amount of heat released per mole of zinc, we get:

`\Delta H_(rxn)=\frac{\text{Heat released in the reaction}}{\text{Moles of zinc}}\\\\\Delta H_(rxn)=(-416.2)/(0.00174)=239195.4J/mol=239.2kJ/mol`

Conversion factor: 1 kJ = 1000 J

Hence, the enthalpy change of the reaction is 239.2 kJ/mol