Question

A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data. Zn ( s ) + 2 HBr ( aq ) ⟶ ZnBr 2 ( aq ) + H 2 ( g )

Answer 1

Answer: The standard enthalpy of the reaction is -248.78 kJ/mol

Step-by-step explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

`q=c* \Delta T`

where,

c = heat capacity of calorimeter = 448 J/K

`\Delta T` = change in temperature = 21.1 K

Putting values in above equation, we get:

`q=448J/K* 21.1K=9452.8J`

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

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 = 2.50 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

`\text{Moles of zinc}=(2.50g)/(65.4g/mol)=0.038mol`

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

`\Delta H^o_(rxn)=(q)/(n)`

where,

`q` = amount of heat released = -9452.8 J

n = number of moles = 0.038 moles

`\Delta H^o_(rxn)` = standard enthalpy of the reaction

Putting values in above equation, we get:

`\Delta H^o_(rxn)=(-9452.8J)/(0.038mol)=-248757.9J/mol=-248.78kJ/mol`

Conversion factor used: 1 kJ = 1000 J

Hence, the standard enthalpy of the reaction is -248.78 kJ/mol