[Sis290025]

What is the delta H in kJ/mol when 0.0636 grams of magnesium are added to 25.0 mL of 6.00 M HCl at 18.8 degrees Celsius. The final temperature of the solution is 29.9 Celsius. Assume that the specific heat of the solution is the same as the heat capacity of water. The calorimeter does not absorb heat.

Mg(s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g) REACTION as I wrote

Now (6.00 M)(.025 L) = 0.15 mol HCl

.15 mol HCl8(36.461 g HCl/1 mol HCl) = 5.46915 g HCl

0.0636 g Mg (1 mol Mg/24.305) = 0.0026167 mol Mg

Therefore, Mg is the limiting reagent??

0.0026167 mol Mg(1 mol H2/1 mol mg) =  0.0026167 mol H2


0.0026167 mol H2*(2.016 g H2/1 mol H2) = 0.005275 g H2

So
q = (4.18 J/g*C)*(29.9 - 18.8 C)*(HCl_mass + Mg_mass - H2_mass)

=(4.18 J/g*C)(29.9 - 18.8 C)*(5.469 g HCl+ 0.0636 g Mg - 0.005275 g H2)
= 0.2567 kJ

delta H = -0.2567 kJ/0.0026167 mol Mg = -98.1 kJ/mol ??


Thank you.



 

[mike]

.15 mol HCl8(36.461 g HCl/1 mol HCl) = 5.46915 g HCl

I don't think you need to calculate this.

Therefore, Mg is the limiting reagent??

Yes.

q = (4.18 J/g*C)*(29.9 - 18.8 C)*(HCl_mass + Mg_mass - H2_mass)

Why subtract the mass of H2, I don't get this. Maybe someone can explain it to me but I have not seen this before. Also I think the mass of HCl should be closer to 25g as you are using 25mL of HCl solution. I think you will find that the equation: Q=m.c.dT assumes in these kinds of experiments that you are using pure water (really, dilute HCl) so the enthalpy is basically calculated from the temperature increase of a certain mass of water. In other words, how much energy is required to heat a known mass of water by a known (measured) temperature.

[Sis290025]

On my other enthalpy post, jdurg explained the subtraction of H2 lost. Is this an unnecessary step? What did I do incorrectly in the calculation of the mass to use in the formula and final calculation in kJ/mol? Are there other ways to get the final answer?

Thanks for any replies and help.

[mike]

Well if you were doing this experiment in the lab you would weigh the amount of water in the caolrimeter to get the mass. Or you could measure the volume, and using a density graph and the water temperature convert this to mass. At 18-22C 25mL of dilute HCl is going to weigh about 25grams.

You are already told to assume the specific heat is the same as the specific heat of pure water so making the assumption that the mass is the same is also valid in this experiment. I would therefore say the mass = 25 + 0.0636 (although even the mass of the Mg won't make loads of difference).

You know the temperature change dT and you know the specific heat of water.

q=m.c.dT

You also know the number of moles of Mg used so you can work out the molar enthalpy.

Plus, you know it increased the temperature of the water so must be exothermic.

[Sis290025]

But why is the molality given if you don't really use it? Is it given because the limiting reagent must be decided after comparing the two reactants in mol to find out which mol amt. you have to divide the kJ by??

[mike]

Sure you can use it to double check that Mg is the limiting reagent.

On my other enthalpy post, jdurg explained the subtraction of H2 lost. Is this an unnecessary step?

If you look carefully at your calculations in your other post after subtracting  the mass of H2 (which is relatively small) you then round your answer up anyway which totally negates the subtraction of the H2 anyway (in other words in wouldn't matter whether you subtracted it or not!)

Another reason the molarity is given may be to show that the HCl is relatively dilute so that it's density is similar to pure water (for the assumptions I mentioned above).

[mike]

But why is the molality given if you don't really use it?

Don't always assume you have to use every piece of information in a question either. Sometimes extra information is given to stump students and see if they really know what they are doing Mean huh?

[jdurg]

Whether or not the mass of the H2 is large or not, you should still make it a habit to remember to subtract it from your final mass.  If you need to do further multiplication/division calculations, that 'small mass' could come back to bite you.  So whenever someone is new to this process, as the original poster seems to be, you should never tell them to 'ignore' any part of it.  

[Sis290025]

So the final answer is

(25 g + 0.0636)*(4.184 J/g*C)*(29.9-18. = -1164.013737 J

-1.164014 kJ / 0.0026167 mol Mg = -444.8 =-445 kJ/mol

Right?

[mike]

Whether or not the mass of the H2 is large or not, you should still make it a habit to remember to subtract it from your final mass.  If you need to do further multiplication/division calculations, that 'small mass' could come back to bite you.  So whenever someone is new to this process, as the original poster seems to be, you should never tell them to 'ignore' any part of it.  

So can you explain to me why we make the assumptions that the heat capacity is for pure water and the density of the solution is for pure water? Are we not just measuring the temperature change of a known amount of water and assuming that it is pure (ie contains no HCl, Mg, H2 etc)? Which factor makes the most difference? I always thought that this "coffee-cup" calorimetry was just a convenient way of heating the water by doing the reaction in the water iteself.

[Sis290025]

Er, is this question directed at me?

[mike]

Er, is this question directed at me?

Nope, not specifically, but if you know the answer you can tell me  

[Sis290025]

I don't know, but I was wondering the same thing.  

Why do we assume water's specific heat and density for solutions like HCl and NaCl? Maybe I should do some research . . .

[Donaldson Tan]

given the concentration of HCl and NaCl, the contribution by these solutes to the solution density is insignificant compared to that by the solvent. Hence, the solution density is approximately the same as the solvent density.

The amount of HCl and NaCl is insignificant to the amount of solvent that their contribution to the overall heat capacity of the solution is insignificant. Therefore, the heat capacity of the solution is effectively that for the solvent.

Of course, these assumptions won't holf if our solution is concentrated.