Consider each of the following reactions and decide whether the standard entropy change for the reaction will be, (a) Positive; or (b) Very close to Zero; or (c) Negative.
Matching pairs
N2 (g) + 3 H2 (g) = 2 NH3 (g)
Mg (s) + CO2 (g) = MgO (s) + CO (g)
C6H12O6 (s) + 6 O2 (g) = 6 CO2 (g) + 6 H2O (l)
2 NO2 (g) = N2O4 (g)
BaSO4 (s) + 2 H2O (g) = BaSO4.2H2O (s)
PCl5 (s) = PCl3 (s) + Cl2 (g)
My answers are:
very close to zero
negative ???
positive
negative
negative
positive
i am not sure of the second one ?? u see the entropy of CO is less than CO2 due to the different arrangements however it can be also close to zero what do u guys think ???
You should get a second opinion on these answers but here is my reasoning. Check my thinking.
a. Four moles goes to two moles. I think S decreases.
b. Both Mg and MgO are crystalline solids. About the same. CO and CO2 both gases. About the same at the same T. So S is about zero.
c. solid and gas goes to gas and liquid. More disorder on the right. S positive.
d. decrease in moles again.
e. gas on left and none on the right. S is negative.
f. gas on the right. none on the left. S is positive.
so the first one is wrong too hmmm ! yeah ur reasoning makes sense ... i will ask my chem teacher too
hmmm interesting
To determine the standard entropy change for each reaction, we need to consider the difference in entropy between the reactants and products. Here's how you can evaluate each reaction:
1. N2 (g) + 3 H2 (g) → 2 NH3 (g)
In this reaction, the number of moles decreases from 4 to 2. Since gases generally have higher entropy than solids, the decrease in the number of gaseous moles leads to a negative entropy change. So, your answer of negative is correct.
2. Mg (s) + CO2 (g) → MgO (s) + CO (g)
To determine the entropy change in this reaction, you are correct that you need to consider the difference in entropy between CO2 (g) and CO (g). The entropy of CO2 is higher due to its greater molecular complexity and increased number of possible microstates. However, when CO2 is converted to CO, it loses some complexity and has a lower entropy value. Although it's not a significant difference, the entropy change is still expected to be very close to zero, especially since the reaction involves the transition between solid and gas phases. Therefore, your answer of very close to zero is likely the most appropriate for this reaction.
3. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
The reaction involves the combustion of glucose (C6H12O6) to form carbon dioxide (CO2) and water (H2O). Since the number of gaseous moles increases from 6 (in O2) to 6 (in CO2), the entropy change is expected to be positive. Therefore, your answer of positive is correct.
4. 2 NO2 (g) → N2O4 (g)
In this reaction, the number of moles decreases from 2 to 1. Similar to the first reaction, the decrease in the number of gaseous moles results in a negative entropy change. Therefore, your answer of negative is correct.
5. BaSO4 (s) + 2 H2O (g) → BaSO4·2H2O (s)
In this reaction, a hydrate (BaSO4·2H2O) is formed by the addition of water molecules to BaSO4. The reaction involves a change from gas to solid, so the entropy change is expected to be negative. Therefore, your answer of negative is correct.
6. PCl5 (s) → PCl3 (s) + Cl2 (g)
In this reaction, the solid PCl5 decomposes to form solid PCl3 and gaseous Cl2. Since the number of gaseous moles increases from 0 to 1, the entropy change is expected to be positive. Therefore, your answer of positive is correct.
Overall, your answers seem mostly correct. For the second reaction, the entropy change is expected to be very close to zero due to the minor difference in entropy between CO and CO2.