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.
I worked this earlier for someone. I can look it up if you need it. Or do you just need to ask a question about it?
I solved it numerous times and it says I am getting the wrong answer. Could you help explain how to solve it and see what I am doing wrong please.
Also can you look it up as well thanks. I really appreciate it!
If you will post your work I will look at it.
http://www.jiskha.com/display.cgi?id=1328846087
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!
To find the enthalpy change (ΔH) for this reaction per mole of water produced, you need to use the equation:
ΔH = q / n
Where:
- ΔH is the enthalpy change in joules (J) per mole (mol) of water produced,
- q is the heat absorbed or released by the reaction in joules (J), and
- n is the number of moles of water produced.
To calculate q, you need to determine the change in heat, which can be found using the equation:
q = m × c × ΔT
Where:
- q is the change in heat in joules (J),
- m is the mass of the solution in grams (g),
- c is the specific heat capacity of water in joules per gram per degree Celsius (J/g°C),
- ΔT is the change in temperature in degrees Celsius (°C).
In this case, since the solution has the same density and specific heat as water, we can assume a density of 1 g/mL and a specific heat capacity of 4.184 J/g°C.
First, calculate the mass of the solution:
Mass of solution = volume of solution × density
= (60.0 mL + 60.0 mL) × 1 g/mL
= 120.0 g
Next, calculate the change in heat (q):
q = mass × specific heat × ΔT
= 120.0 g × 4.184 J/g°C × (28.70 °C - 24.07 °C)
Then, calculate the number of moles of water produced. Since the reaction is between Ba(OH)2 and HCl, it produces 2 moles of water per mole of Ba(OH)2 reacted:
Number of moles of water = 2 × (0.0600 L) × (0.340 mol/L)
= 0.0408 mol
Finally, calculate the enthalpy change (ΔH) per mole of water produced:
ΔH = q / n
= (calculated q value) / 0.0408 mol
By substituting the calculated values into the equation, you can find the enthalpy change per mole of water produced.