[Sis290025]

When 1.50 g of Ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 C to 33.10 C. If the specific heat of the solution is  4.18 J/g*C, calculate  for the reaction, as written.
 Ba(s) + 2H2O --> Ba(OH)2 + H2 (g) deltaH = _?__
 
a.431 kJ  
 
b.-3.14 kJ  
 
c.-431 kJ    
 
d.3.14 kJ  
 
When I attempted to do this problem my answer came out to -425 kJ, and I wanted to know if my steps are completely off.

quantity of heat absorbed by H2O = (100 g)*(4.18 j/g*C)*(33.1 - 22 C) = 4639.8 J
q_rxn = -4.639 kJ

delta H = -4639.8 J/(1.5 g Ba/137.33 g Ba) = 4.23787 kJ
What is the proper way of doing this problem?

Thanks.

[jdurg]

Remember that the water you have is a reactant in the equation, and while you gain some mass of the added barium, you lose some mass from the escaping hydrogen gas.

So what you need to do is calculate what your total mass of reactants is.  (Mass of H2O and Ba).  Then you need to calculate the mass of hydrogen that is lost and subtract that from your previous total.  This will give you the mass of the final product (The Ba(OH)₂ solution) which is what you have the specific heat value for.  At this point you plug in the value that you calculated in place of the (100 g) in your equation below and you will come up with the right answer.  

[Sis290025]

The total mass in g of the reactants is is the mass of Ba (1.50 g) + mass H2O (100 g) = 101.5 g? To calculate the g of H2, I set up a molar ratio with Ba, which is the limiting reagent?

[jdurg]

The total mass in g of the reactants is is the mass of Ba (1.50 g) + mass H2O (100 g) = 101.5 g? To calculate the g of H2, I set up a molar ratio with Ba, which is the limiting reagent?

Yes.  The total mass of the reactants is 101.50 grams.  To calculate the mass of H2 produced you are correct to set up a molar ratio.  To find a limiting reactant, just calculate the number of moles of Ba you have and the number of moles of H2O you have.  You can see that one mole of Ba will make one mole of H2 gas.  So how many moles of Ba is 1.50 grams?  How much is the mass of that number of moles of H2 gas?  (Remember that it's H2 gas which is twice the molecular mass of an H atom).

[Sis290025]

My workings:

1.50 g Ba*(1 mol Ba/137.33 g Ba) = 0.01092250 mol Ba

100g H2O (1 mol H2O/18.016 g H2O) = 5.55062 mol H2O

If one looks at the H2O:Ba ratio, H20 is in excess, so Ba is limiting reagent.

0.01092250 mol Ba (1 mol H2/1 mol Ba) = 0.01092250 mol H2

0.01092250 mol H2 ( 2.016 g H2/1 mol H2) = 0.02201996 g H2

So mass of mixture is 101.5 g - 0.02201996 g H2 = 101.47798 g


101.47798 g(4.18 J/g*C)(33.1 -22 C) = 4708.3753 J = -4.71 kJ  :aak:

What am I doing incorrectly? Am I at least correct in assuming the final answer is negative?

Thanks again.  

[mike]

-4.71 kJ  

not incorrect, you just need to divide by the number of moles of Ba to get your answer in kJ.mol^-1

[Sis290025]

Is the delta H normally expressed in kJ/mol? Why do the choices only have kJ as the units?

So the answer is -431 kJ/mol?

Thanks again.  

[mike]

I gues it depends on the question being asked. If you were asked for the amount of energy released by one mole of reactant then the answer would be in kJ. If you are asked for a molar enthalpy change then kJ.mol-1.

I don't know why your choices only have kJ and not kJ.mol-1 did you write the question verbatim or did you change it?

[Sis290025]

I checked, and it undoubtably had kJ and not kJ/mol.  

Thanks for tip on the units.