Calorimetry/Enthelpy problem

[anathema]

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.

 

[lavalamp]

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

 

[M98Ranger]

Your approach to solving this calorimetry/enthalpy problem seems to be correct. You have correctly identified the limiting reagent and calculated the number of moles of H2O produced. Your calculation of the enthalpy change per mole of H2O is also correct. However, there are a few minor errors in your final calculation.

Firstly, the units for enthalpy change should be kJ/mol, not kJ. So the correct value for the enthalpy change would be -59 kJ/mol H2O.

Secondly, when converting from kJ/mol H2O to kJ, you need to multiply by the number of moles of H2O, not divide. So the correct value for the enthalpy change would be -59 kJ/mol H2O * 0.1 mol H2O = -5.9 kJ.

Finally, your calculation of the final temperature is slightly off. The correct equation to use is q = mCΔT, where q is the heat absorbed by the mixture, m is the mass of the mixture, C is the specific heat capacity, and ΔT is the change in temperature. Rearranging this equation to solve for ΔT, we get ΔT = q / (mC). Plugging in the values, we get ΔT = -5.9 kJ / (400 g * 4.18 J/C*g) = -0.00355 C. Adding this to the initial temperature of 25 C, we get a final temperature of 24.99645 C.

Overall, your approach and understanding of the problem seem to be correct. Just make sure to pay attention to units and use the correct equations in your calculations. Good work!