I know I posted this question again but no one is really helping me. Can someone please help me as soon as possible.
In a constant-pressure calorimeter, 60.0 mL of 0.340 M Ba(OH)2 was added to 60.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 24.07 °C to 28.70 °C. If the solution has the same density and specific heat as water, what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
This is my work.....
(18g) (4.184 J)(28.70-24.07)
I get 350.95
then i divide it by 0.0340 and get 10322.17. Then I divide it by a 1000 and get 10.322... It is wrong. Can you take a look at my work and tell me what i did wrong and confirm the right answer for me thanks!
Where do you come up with 18g? That is 60 mL of one reagent plus 60 ml of the other. With a density of 1.00 g/mL, the water has a mass of 120 grams. So substitute 120 for the 18 and recalculate that part.
The second error is in the moles. I calculated 60.0 mL x 0.340M = 20.4 mmoles Ba(OH)2 and
60.0 mL x 0.680M = 40.8 mmoles HCl. Therefore mol H2O formed is
......Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
......0.0204...0.0408...0.0204.0.0408
So q from above divided by 0.0408 gives q/mole H2O. Watch the s.f., a common problem with these databases.
Your post doesn't give if that is to be J/mol or kJ/mol. Watch that too.
I got an answer of 56.976. The answer is negative though since the reaction is exothermic. Other than that great explanation. Thank you so much. You are the best!!!! :]
yeah i converted it. It was suppossed to be kj :]
To find the enthalpy change (ΔH) for a reaction using a constant-pressure calorimeter, you can use the equation:
ΔH = q / n
Where:
ΔH is the enthalpy change,
q is the heat absorbed or released by the reaction, and
n is the number of moles of the substance involved in the reaction.
To find the heat absorbed or released by the reaction (q), you can use the equation:
q = m × c × ΔT
Where:
q is the heat absorbed or released,
m is the mass of the solution in grams,
c is the specific heat capacity of the solution (assumed to be the same as water, which is 4.18 J/g°C),
and ΔT is the change in temperature (final temperature - initial temperature).
First, let's calculate the mass (m) of the solution:
Since the density and volume of the solution are not given, we assume the density and specific heat of water, which are 1 g/mL and 4.18 J/g°C, respectively.
Using the volume of the solution, we can calculate the mass (m) of the solution:
mass (m) = volume × density = (60.0 mL + 60.0 mL) × 1 g/mL = 120 g
Next, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature = 28.70 °C - 24.07 °C = 4.63 °C
Now, let's calculate the heat absorbed or released by the reaction (q):
q = m × c × ΔT = 120 g × 4.18 J/g°C × 4.63 °C = 2285.77 J
Finally, let's calculate the number of moles of water produced in the reaction:
The balanced equation for the reaction is:
Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l)
From the balanced equation, we can see that 2 moles of H2O are produced for every 1 mole of Ba(OH)2 reacted.
The volume of Ba(OH)2 solution is given as 60.0 mL, which can be converted to liters:
volume = 60.0 mL × (1 L / 1000 mL) = 0.06 L
Using the volume and concentration of the Ba(OH)2 solution, we can calculate the number of moles (n) of Ba(OH)2:
n = volume × concentration = 0.06 L × 0.340 mol/L = 0.0204 mol
Since 2 moles of H2O are produced per 1 mole of Ba(OH)2, the number of moles of water (n) produced is twice the number of moles of Ba(OH)2:
n(H2O) = 2 × 0.0204 mol = 0.0408 mol
Now, we can calculate the enthalpy change (ΔH) for the reaction per mole of H2O produced:
ΔH = q / n = 2285.77 J / 0.0408 mol = 55991.8 J/mol
Therefore, the enthalpy change (ΔH) for this reaction per mole of H2O produced is approximately 55991.8 J/mol.