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Calorimetry/Enthalpy problem

  • Thread starter anathema
  • Start date
5
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This problem's been confusing the heck out of me.
"Consider the reaction:
2HCl(aq) + Ba(OH)2(aq) -> BaCl2(aq)+2H2O(l) deltaH= -118 kJ

Calculate the heat when 100.0 mL of 0.5 M HCl is mixed with 300.0 mL of 0.5 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25 C and that the final mixture has a mass of 400.0 g and a specific heat capacity of 4.18 J/C*g, calculate the final temperature of the mixture."

There is no answer given to me for this problem so I want to make sure if I am doing this right, which I doubt I am.
The net reaction is H2 + 2(OH) -> 2H2O deltaH= -118

The reason I chose H2 as opposed to 2H is because that's how H occurs in its natural state, and it would still respect the initial molar quantities. Now I found the limiting reagent. I know I have 0.05 mols of H2 and .15 mols OH, and if I divide .15 by 2mols OH then .075 > .05 so H2 is the limiting reagent. I then find out how many mols of H2O this would give me, which would be 0.05 * 2/1 = 0.1 mols H2O. Now I find the kJ/mol H2O enthalpy, which is -118/2 = -59 kJ/mol H2O. Multiplying that by .1 gives me -5.9 kJ. Since the reaction is losing energy, the energy perceived by the calorimeter is positive and so there will be an increase in temperature. So 5900 J = 4.18 * 400 * (x-25), x = 28.52871 C. Is this right? Thanks.
 
279
1
Well it gives you the [del]Hmol-1 for the reaction as -118 KJmol-1, and from the equation you can see that 2 moles of HCl react with Ba(OH)2, all you gave to do to find the energy change is to find out how many moles of each you have in the question.

moles of HCL = c * v = 0.5 * 0.1 = 0.05 moles

moles of Ba(OH)2 = c * v = 0.5 * 0.3 = 0.15 moles

Since the ratio is 2:1, you can see that there is an excess of Ba(OH)2, therefore it will be 0.05 moles of HCl reacting with 0.025 moles of Ba(OH)2.

[del]Hmol-1 for reaction = -118 KJmol-1
[4] energy absorbed = 0.05/2 * -118 = -2.95 KJ

[4] energy released = 2.95 KJ

Since E = c * m * T

2950 = 4.18 * 400 * [del]T
[del]T = 1.67 K

[4] temperature = 25 + 1.67 = 26.7 C
 

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