Suppose that 50.0 mL of a 0.250 M HCl solution is mixed with 50.0 mL of a 0.100 M
Ba(OH)2 solution in a coffee-cup calorimeter. The temperature of the resulting solution increases by
1.76 oC . Calculate DH for the reaction between HCl (aq) and Ba(OH)2 (aq) in kJ per mole of Ba(OH)2 reacted. Assume that the specific heat capacity of the resulting solution is 4.18 J/g oC and
that the density of the resulting solution is 1.00 g/mL . Also assume that the heat capacity of the
calorimeter itself can be ignored.
If you expect to continue getting help you should show some work. Or something about how you would attack the problem. Or EXACTLY what you don't understand about the process.
q = [mass H2O x specific heat H2O x delta T]
Then dH is q/mol Ba(OH)2 and convert to kJ.
175j
To calculate ΔH for the reaction between HCl and Ba(OH)2, we can use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat transferred during the reaction, and n is the number of moles of Ba(OH)2 reacted.
First, let's calculate the amount in moles of Ba(OH)2 reacted:
n = C * V
where n is the number of moles, C is the concentration in moles per liter (M), and V is the volume in liters.
For Ba(OH)2 solution:
C = 0.100 M
V = 50.0 mL = 0.0500 L
n = (0.100 M) * (0.0500 L) = 0.00500 moles
Next, we need to calculate the heat transferred during the reaction (q). We can use the formula:
q = mcΔT
where q is the heat transferred, m is the mass of the resulting solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, let's calculate the mass of the resulting solution:
mass = density * volume
mass = (1.00 g/mL) * (50.0 mL + 50.0 mL) = 100.0 g
Now let's calculate the heat transferred:
q = (100.0 g) * (4.18 J/g oC) * (1.76 oC) = 736.08 J
Finally, let's calculate ΔH:
ΔH = q / n
ΔH = (736.08 J) / (0.00500 mol) = 147,216 J/mol
To convert J to kJ, divide by 1000:
ΔH = 147,216 J/mol / 1000 = 147.22 kJ/mol
Therefore, the ΔH for the reaction between HCl and Ba(OH)2 is 147.22 kJ/mol of Ba(OH)2 reacted.