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. Using these data, calculate the standard enthalpy of reaction.Zn(s) + 2HBr(aq) --> ZnBr2(aq) + H2(g)

Answer 1

247 kJ/mol

Step-by-step explanation:

Zn(s) + 2HBr(aq) → ZnBr₂(aq) + H₂(g)

First, we need to find the limiting reactant. And to do that, we need to find the amount of moles of each reactant.

2.50 g Zn * (1 mol / 65.38 g) = 0.03824 mol Zn

0.1000 L * 2.00 mol/L = 0.200 mol HBr

Since 1 mol of Zn reacts with 2 mol of HBr, it is clear that Zn is the limiting reactant.

The amount of heat can be calculated as:

q = (448 J/K) * (21.1 K)

q = 9452.8 J

So the standard enthalpy of reaction is:

ΔH = (9452.8 J) / (0.03824 mol Zn)

ΔH = 247 kJ/mol