Example Problems-Thermochemistry
1. If a 40.1 g piece of iron at 652 °C is dropped into a sample of 328 g of water at 32.4 °C, what will the final temperature
be after thermal equilibrium is established? Assume that no heat is lost during the process. (Obtain specific heat
capacities from your text.)
2. (a) How much heat (in kJ) is evolved or absorbed in the reaction of 233.0 g of carbon with enough CaO to produce
calcium carbide? (b) Is the process exothermic or endothermic?
CaO(s) + 3 C(s)  CaC2(s) + CO(g);
H = 464.8 kJ
3. 85.8 kJ of energy is evolved (i.e., released) at constant pressure when 3.56 g of P4 is burned according to:
5 O2(g)  P4O10(s) What is the H for the (thermo)chemical equation?
P4(s) +
4. Instant cold packs contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and
the solid dissolves, lowering the temperature because of the endothermic process:
NH4NO3(s)  NH4NO3(aq); H = +25.7 kJ
What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 125 mL of water?
Assume the specific heat capacity of the dissolved NH4NO3 is negligible compared to water, an initial temperature of
25.0 C, and no heat transfer between the cold pack and the environment. (Recall, dwater ~ 1.0 g/mL)
5. Assume that a particular reaction carried out at constant pressure produces 244 kJ of heat and that 35 kJ of PV
(expansion/contraction) work is done on the system. What are the values of E and H for the system? For the
surroundings?
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Thermochemistry Practice Problems (continued)
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 = ???
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)
8. Calculate H for the following equation:
PbCl2(s) + Cl2(g)
Given:
Pb(s) + Cl2(g)
Pb(s)
+ 2 Cl2(g)
+ H2O;
 PbCl4(l)
 PbCl2(s)
 PbCl4(l)
H = ???
H = ???
H1 = -359.4 kJ
H2 = -329.3 kJ
9. Calculate H for the following equation:
2 C(s) + 3 H2(g)  C2H6(g);
H = ???
given the following data:
2 C2H6(g) + 7 O2(g)
 4 CO2(g)
C(s) + O2(g)
2 H2(g)
+ O2(g)
+ 6 H2O(l);
 CO2(g) ;
 2 H2O(l);
2
H = -3120 kJ
H = -394 kJ
H = -572 kJ
Prof. Mines’s Strategy for approaching “Hess’s Law” type problems
What I’m calling a “Hess’s Law” problem is one in which you are given a “target” chemical equation whose H
you would like to figure out, and you are given two or more other chemical equations with their H values that
you are to “use” in order to find the H value you want. An example is:
Target equation
Given the data below, calculate the H for:
Given equations
(my terminology)
ClF(g) + F2(g)
→ ClF3(g)
(1) 2 ClF(g) + O2(g) → Cl2O(g) + F2O(g)
H1 = 167.4 kJ
(2) 2 ClF3(g) + 2 O2(g) → Cl2O(g) + 3 F2O(g)
H2 = 341.4 kJ
(3) 2 F2(g) + O2(g) → 2 F2O(g)
H3 = - 43.4 kJ
(my terminology)
The general strategy is as follows. You will generate a series of chemical equations one by one, which when
added together will exactly match the equation whose H you are trying to determine (the target equation). If
you use variants of the equations whose H values are known/given in the problem (given equations), you will
“know” the H value for each of the equations in your series. If you add up all of these H’s, the result will equal
the H you seek. Why? Because H is a state function, which means that the H between two “states” does not
depend on how you get from state 1 to state 2. This means that the Hoverall = Ha + Hb + …. as long as the
sum of processes a, b, etc. equals the overall process (Hess’s Law). So, how do you generate the series?
1) Consider the first substance in the target equation, and look for that substance in one of the given
equations. Note that you might find this substance as either a product or a reactant in one (or more) of
the given equations. If the substance is present in more than one of the given equations, then
ignore that subtance and start this procedure over with the next substance in the target
equation. If the substance being considered is present in only one equation (or one remaining equation;
see #5 below), consider that given equation and proceed to step 2. NOTE: If the substance of interest
only appears in a crossed out equation (see #5 below), then ignore that substance and go on to the next
substance in the target equation.
2) If the coefficient for the substance of interest in the given equation is not the same as its coefficient in
the target equation, multiply the given equation (i.e., all coefficients) by whatever number is needed to
get the number to match the coefficient in the target equation.
3) If the substance of interest is on the opposite side in the given equation as the side in which it appears
in the target equation, reverse the given equation (put what’s on the right side on the left and vice
versa). After this “flipping”, the substance should be on the same side in the reversed equation and the
target equation. I’ll call the equation that you’ve generated in Steps 2 and 3 a “modified given equation”.
4) Determine the H for the modified given equation you created in Steps 2 and 3 by doing the following:
i) multiply the H for the given equation by the same number you multiplied the coefficients by in Step
2 (if applicable), and
ii) multiply the H by -1 (if the process was reversed in Step 3). NOTE: This does not necessarily
mean that the H will be negative! You are changing the sign of H (from either + to – OR – to +).
5) Cross out the given equation that you just used in Steps 2 and 3. You should not use this equation
again in this procedure! Only consider the “unused” or “remaining” given equations from here on out!
6) Look at the next substance in the target equation and do the same as in (1) - (4) above, being careful
not to reuse any given equations (i.e., use only “remaining” given equations; not crossed out ones).
7) Continue on until you have reached the last substance in the target equation.
8) Sum up all of the “modified given equations” you’ve generated and make sure the sum matches the
target equation. If not, look for an unused given equation that can be made to “cancel out” any
unwanted substances in your current sum. (Very often things “work out” if you just go substance by
substance through the target equation as I have instructed above, but occasionally you may find that
you need to “fiddle” to get everything to sum up correctly.)
9) Sum up the H’s for all the modified given equations to get the overall H!
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