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. Using these data, calculate the standard enthalpy of reaction.
Zn(s)+2HBr(aq) --> ZnBr2(aq)+H2(g)
To calculate the standard enthalpy of reaction, we need to use the equation:
ΔH° = q / n
where ΔH° is the standard enthalpy of reaction, q is the heat transferred during the reaction, and n is the number of moles of the limiting reactant.
To solve this problem, we need to determine the heat transferred during the reaction and the number of moles of the limiting reactant. Let's break down the steps:
Step 1: Calculate the heat transferred (q)
In a calorimeter, the heat transferred (q) is given by the formula:
q = CΔT
where q is the heat transferred, C is the total heat capacity of the calorimeter and solution, and ΔT is the change in temperature.
In this case, we are given that C = 448 J/K and ΔT = 21.1 K, so we can calculate q as follows:
q = 448 J/K * 21.1 K
q = 9444.8 J
Step 2: Calculate the number of moles of the limiting reactant (n)
To determine the limiting reactant, we need to compare the stoichiometric ratios from the balanced chemical equation.
From the balanced equation: Zn(s) + 2HBr(aq) -> ZnBr2(aq) + H2(g)
We can see that the stoichiometric ratio between Zn and HBr is 1:2. Given that we have 2.50 g of Zn, we need to convert it to moles.
First, calculate the molar mass of Zn:
Molar mass of Zn = 65.38 g/mol
Now, calculate the number of moles of Zn:
moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 2.50 g / 65.38 g/mol
moles of Zn = 0.0382 mol
Since the stoichiometric ratio between Zn and HBr is 1:2, the number of moles of HBr is twice that of Zn:
moles of HBr = 2 * 0.0382 mol
moles of HBr = 0.0764 mol
Step 3: Calculate the standard enthalpy of reaction (ΔH°)
Using the equation ΔH° = q / n, we can now calculate the standard enthalpy of reaction:
ΔH° = 9444.8 J / 0.0764 mol
ΔH° = 123547.6 J/mol
Since enthalpy is typically reported in kilojoules per mole (kJ/mol), we can convert the result to kJ/mol by dividing by 1000:
ΔH° = 123.55 kJ/mol
Therefore, the standard enthalpy of reaction for the given reaction is 123.55 kJ/mol.