Answers, Thermochemistry Practice Problems 2
6. When 26.7 g of H2S was burned in excess oxygen, 406 kJ was released. What is H for the following
equation?
2 H2S(g) + 3 O2(g)
2 SO2(g) + 2 H2O(g);
H = ???
Answer: -1040 kJ
Reasoning:
26.7g
34.08 g/mol H2S
0.78345 mol H2S , so 406 kJ was released when 0.78345 moles of H2S reacted.
406 kJ is released
0.78345 mol H2S that reacts
518.2 kJ released/mol H2S that reacts
Since the coefficient of H2S is a “2” here, you need to find out the energy change associated with TWO
moles of H2S reacting:
518.2 kJ is released
(1) mol H2S that reacts
2 mol H2S that reacts
1036
1040 kJ is released
Add the negative sign because of the energy being RELEASED.
Thus, H = -1040 kJ for the thermochemical equation above.
7. You mix 200.0 mL of 0.200 M RbOH(aq) with 100.0 mL of 0.400 M HBr(aq) in a coffee cup
calorimeter. If the temperature of each of the two solutions was 24.40 °C before mixing, and the
temperature raised to 26.18 °C, what is the H associated with the thermochemical equation?
RbOH(aq) + HBr(aq) → RbBr(aq) + H2O;
H = ???
Answer: -55.9 kJ
Reasoning:
As in problem #1 above, there is a “process” occurring in the system. So the temperature of the
water increased as a result of that process (not because of an initial difference in T between the
system and the surroundings) and we can say (choosing the atoms/ions in all substances involved in the reaction
as the “system”):
Hsys= - qwater
Hsys
= -q water =
4.184 J
26.18 °C - 24.40 °C (200. + 100. g) = -
g × °C
4.184 J
g × °C
(1.78 °C)(300. g) = -2234 J
So, Hsys =-2234 J = -2.23 kJ for this actual reaction (not for the “equation”)
Notes: This problem is essentially identical to the calculations you did in lab with your data from reacting HCl with
NaOH (Expt. 14, part C)!! As in the lab, to estimate an answer to this problem using the information given, one
must assume that the density of the solutions is 1.00 g./mL and that the specific heat of the solution that is formed is
the same as that of water. That is, we basically consider the energy required to heat the newly formed products to
be insignificant compared to the energy needed to heat the water. Note that the number of significant figures in T
becomes THREE because of the subtraction. In your experiment, this will likely drop to two since our thermometers
measure to the tenths place, not the hundredths.
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Answers, Thermochemistry Practice Problems 2
The “complete” thermochemical equation is:
RbOH(aq) + HBr(aq) → RbBr(aq) + H2O;
H = ???
The H value appropriate for the thermochemical equation is the one that corresponds to one
mole of RbOH and one mole of HBr reacting to form one mole of H2O (because those amounts
correspond to the coefficients in the balanced equation).
So we must determine how much reaction occurred when we mixed the two solutions that yielded the
2.23 kJ of energy that was released:
0.200 mol RbOH
mol RbOH =
mol HBr =
L
0.400 mol HBr
L
0.2000 L of solution
0.1000 L of solution
0.0400 mol RbOH
0.0400 mol HBr
So there is no “limiting reactant” here; all of both react. So we need to “scale up” the amount of
reaction from 0.0400 mol to “one mole” (using the “tuna can” idea again (price per ounce)):
-2.234 kJ
H for 1 mol of RbOH reacting
-55.85
0.0400 mol
RbOH(aq) + HBr(aq) → RbBr(aq) + H2O;
-55.9 kJ / mol
H = -55.9 kJ
NOTE: Since RbOH and RbBr are soluble, they are strong electrolytes. And HBr is one of the six
+
-
strong acids, so it also is a strong electrolyte. Thus Rb and Br will be spectator ions and the net
ionic (thermochemical) equation is just:
OH-(aq) + H+(aq)
H2O(l)
This net ionic equation describes identically the reaction that you did in Part C of Experiment 14! (Verify that for
yourself!) How did your value compare to this (presumably more accurate) result?
8. Calculate H for the following equation:
PbCl2(s) + Cl2(g)
Given:
PbCl4(l)
Pb(s) + Cl2(g)
Pb(s) + 2 Cl2(g)
H = ???
PbCl2(s)
H1 = -359.4 kJ
PbCl4(l)
H2 = -329.3 kJ
Answer: +30.1 kJ
Reasoning: You need one mole of PbCl 2(s) as a reactant, so reverse the first equation (and don’t
forget to reverse the sign of H). You need one mole of PbCl 4(l) as a product, so keep the second
equation as is. The “sum” of these two processes yields the equation (process) desired.
PbCl2(s)
Pb(s) + Cl2(g)
Pb(s) + 2 Cl2(g)
PbCl2(s) + Cl2(g)
PbCl4(l)
PbCl4(l)
H1 ’ = +359.4 kJ
H2 ’ = -329.3 kJ
H = 359.4 + (–329.3) = +30.1 kJ
This is a lot easier problem than #9 which I did on the board. Make sure you can do more complicated
versions of these. Note that if you needed 3 moles of PbCl2(s) for example, you would need to multiply
the entire thermochemical equation by 3 (including the H!). Also, try to identify reactants or products that
are found ONLY in ONE equation that is given and deal with them first. That should make things much
more straightforward. See handout I gave on a “procedure” for doing Hess’s Law type problems.
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Answers, Thermochemistry Practice Problems 2
9. Calculate H for the following equation:
2 C(s) + 3 H2(g)
C2H6(g);
H = ???
given the following data:
(1)
2 C2H6(g) + 7 O2(g)
(2)
C(s) + O2(g)
(3)
2 H2(g) + O2(g)
4 CO2(g) + 6 H2O(l);
CO2(g);
2 H2O(l);
H1 = -3120 kJ
H2 = -394 kJ
H3 = -572 kJ
Answer: -86 kJ
Reasoning: I’ll give you the short version here (since I [hopefully!] did this on the board and I’m
rushed at the moment):
NOTE: I numbered the equations above (1, 2, and 3) to make it easier to keep track of which equation (and H)
I’m using where. I suggest you do the same when you work such problems.
1) Need 2 C(s) on left => Take equation (2) and multiply it by 2
2) Need 3 H2(g) on left => Take equation (3) and multiply it by 3/2
3) Need 1 C2H6(g) on right => Take equation (1) and reverse it as well as multiplying it by 1/2
If you sum all three of the resulting equations, you will get the overall target equation desired! (check
that yourself) By Hess’s Law, the Hoverall will become H2’ + H3’ + H1’ where:
H2’ = 2 x (-394 kJ) = -788 kJ;
H3’ = 3/2 x (-572) = -858 kJ; and
H1’ = -1 x 1/2 x (-3120) = +1560 kJ (note the change in sign for this one!)
Hoverall = -788 + (-858) + 1560 = -86 kJ
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