Most of the question came out in the preliminary examinations or the A levels H2 Chemistry examinations usually involve standard definitions that most students are familiar with…
such as enthalpy change of solution, formation, combustion, etc. However, once in a while, there are a few questions that are totally different, and as such, many students will be thrown off their comfort zones and have problems constructing energy cycles.
An example is found in VJC 2012, paper 3:
Below are some thermodynamic equations involving zinc compounds.
|ZnO(s) + 3C(s) → ZnC2 (s) + CO (g) …… (1)||ΔHθ = +462 kJmol-1|
|ZnO(s) + H2O (l) → Zn(OH)2 (s) …… (2)||ΔHθ = -65kJmol-1|
|ZnC2(s) + 2H2O(l) → Zn(OH)2 (s) + C2H2 (g) … (3)||ΔHθ = -126 kJmol-1|
2 H2O (l) → 2H2(g) + O2(g) …… (5)
ΔHθ = +572 kJmol-1
|Using these data together with the following data,|
|2C(s) + O2 (g) → 2CO2 (g) …… (4)||ΔHθ = -221 kJmol-1|
Determine the standard heat of formation of C2H2 (g) with the aid of an energy cycle.
This equation is a rehash of a much older question that appeared in the early 2000s
Step 1 :
Know what is the unknown in this question
|2C(s) + H2 (g)||→||C2H2(g)||……….. (*)|
Step 2 :
Search, amongst the data, for equation (s) that contain (s) either the left or right hand side of the equation for the unknown enthalpy.
Notice that the equation (3) in the data given contains C2H2 on the right side. C and H2 in the unknown are not found in any single equations in the data given.
Thus, it would be wise to start with the equation (3) when constructing the energy cycle.
|ZnC2(s) + 2H2O(l)||→||Zn(OH)2 (s) + C2H2 (g)||……… (3)|
Now that we have the right hand side of (*) in place, how do we made use of the other data to form the cycle? Remember that we need to have the left hand side of (*) in the cycle too, for the unknown ΔH to be formed.
In the equation (3) above, notice that there are 3 compounds that are not wanted in (*), ie ZnC2, H2O and Zn(OH)2.
Let’s attempt to remove one of the unwanted compounds in (1) above, for example ZnC2.
Notice that ZnC2 is found on the right hand side of (1). But CO is also found on the right hand side of (1). Thus, we will attempt to add CO to the left and right hand side of (3), so that ZnC2 can react with CO. Adding CO on the right hand side of (3) is to make sure that equation (3) is still balanced.
|CO (g) + ZnC2(s) + 2H2O(l)||→||Zn(OH)2 (s) + C2H2 (g) + CO(g)||……… (3)|
Step 3 : Connect equation (1) with (3)
Step 4: Bring in equation (2)
Step 5: Bring in equation (5)
Step 6: Bring in equation (4)
Step 7 : Notice that the final gap is the reaction whose enthalpy we are supposed to find? Now we just need to link them together!
With the cycle constructed, 70-80% of the marks is in the pocket. The final step is to put the enthalpy changes together to form an equation and solve for the unknown
This example shows us that in some cases, the addition of compounds might be necessary for the completion of the cycle / energy level diagram. One thing to remember is that whatever compounds that are added to one side of the equation needs to be added in equal moles to the other side, or the equation will be out of balance.
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