In a constant-pressure calorimeter, 50.0 mL of 0.300 M Ba(OH)2 was added to 50.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 24.50 degrees C to 28.59 degrees C. If the solution has the same density and specific heat as water, what is delta H for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes. Delta H = ___ kJ/mol H2O

Question

In a constant-pressure calorimeter, 50.0 mL of 0.300 M Ba(OH)2 was added to 50.0 mL of 0.600 M HCl. The reaction caused the temperature of the solution to rise from 24.50 degrees C to 28.59 degrees C. If the solution has the same density and specific heat as water, what is delta H for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes. Delta H = ___ kJ/mol H2O

Answer 1

See your earlier post. And you responded to it. Why are you posting the same problem again?

Answer 2

To find ΔH for the reaction (per mole of H2O produced), we need to use the equation:

ΔH = q / n

where ΔH is the enthalpy change, q is the heat absorbed or released, and n is the number of moles of H2O produced.

To find the heat absorbed or released (q), we can use the equation:

q = m * C * ΔT

where m is the mass, C is the specific heat, and ΔT is the change in temperature.

First, we need to calculate the mass of the solution. Since the solution has the same density and specific heat as water, we can assume that the density of water is 1 g/mL. Therefore, the mass of the solution is:

mass = volume * density
= (50.0 mL + 50.0 mL) * 1 g/mL
= 100.0 g

Next, we can calculate the change in temperature (ΔT) using the initial and final temperatures:

ΔT = final temperature - initial temperature
= 28.59°C - 24.50°C
= 4.09°C

Now, we can calculate the heat absorbed or released:

q = m * C * ΔT
= 100.0 g * 4.18 J/g°C * 4.09°C
= 1713.72 J

Next, we need to find the number of moles of H2O produced. From the balanced equation:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

We can see that 2 moles of H2O are produced for every mole of Ba(OH)2 used. Therefore, the number of moles of H2O produced is equal to the number of moles of Ba(OH)2 used:

moles of H2O = moles of Ba(OH)2
= volume of Ba(OH)2 * molarity of Ba(OH)2

The volume of Ba(OH)2 used is given as 50.0 mL. Thus:

moles of H2O = 0.050 L * 0.300 mol/L
= 0.015 mol

Now, we can calculate ΔH using the equation:

ΔH = q / n
= 1713.72 J / 0.015 mol
= 114,248 J/mol

Finally, we need to convert the answer from Joules to kilojoules by dividing by 1000:

ΔH = 114,248 J/mol / 1000
= 114.248 kJ/mol

Therefore, ΔH for this reaction (per mole of H2O produced) is 114.248 kJ/mol H2O.