5.1 – Practice Assignment
1.
Write reversible reactions for each of the following situations (be sure to balance your equations):
a. Hydrogen iodide gas (HI) decomposes into its elements.
2HI (g) ↔ H2 (g) + I2 (g)
b.
2.
Hydrogen and nitrogen gases combine to form ammonia gas, NH3.
3H2 (g) + N2 (g) ↔ 2NH3 (g)
If the system represented by the following equation is found to be at
equilibrium at a specific temperature, which of the following statements is
true? Explain your answers.
H2O(g) + CO(g) ↔ H2 (g) + CO2 (g)
a. All species must be present in the same concentration.
b. The rate of the forward reaction equals the rate of the reverse reaction.
c. We can measure continual changes in the reactant concentrations.
a) False; at equilibrium, total amount of particles will not change, but the concentrations are not
necessarily the same/consistent for all of the reaction participants – they can fluctuate
b) True; this is the definition of equilibrium
c) False; since the forward rate of reaction is equal to the reverse rate of reaction, concentrations of the
reaction participants will increase and decrease accordingly; overall there is no net change in
concentrations of total reaction participants
3.
-
Would you expect the combustion of methane, CH 4 with oxygen to form
carbon dioxide and water, to be a reversible reaction?
Hint: Methane, or natural gas, is an important energy source. Considering
this, what did you learn in the thermodynamics unit that will help you predict whether or not the reverse
reaction is likely to be spontaneous?
1CH4 (g) + O2 (g) ↔ 1CO2 (g) + 2H2O (g)
Not reversible because the forward reaction releases a LARGE amount of energy. Recall the Hess’s Law
Equation: ΔH = ΣΔHf (products) - ΣΔHf (reactants) when we calculate the ∆H we get a value of 890.3kJ. This is highly exothermic, therefore the reverse reaction would require a lot of energy to take
place. Therefore it would not be a reversible reaction and it would not be spontaneous either.
4.
5.
This system is NOT at equilibrium because the system is OPEN. Gas is escaping from the liquid, but no
gas is going back into the liquid
Yes, the system is at equilibrium. Water is evaporating in the flask and vapour is re-entering the liquid
at the same rate
6.
This will not produce pure COCl2 (g) because there will always be some CO (g) and Cl2 (g) in equilibrium
with the COCl2 (g). If some pure COCl2 could be obtained, it would partly decompose into CO (g) and Cl2
(g) as equilibrium occurs
7.
A) The gases inside the tube are the same color – “medium red-brown”
b) Yes; the color in the tubes changes when different temperatures are applied. This shows that
different substances were predominantly present at different temperatures
c) The color change stops and remains constant; if the temperature were increased further, the color
would potentially get darker
d) This reaction would be ENDOTHERMIC:
N2O4 (g) ↔ 2NO2 (g)
(clear)
(red/brown)
When we add heat to the system we produce more NO2. What this means that:
N2O4 (g) + heat ↔ 2NO2 (g) .
When we add heat to the system on the forward reactants side, we drive the reaction to the right,
producing more NO2
e) Low = N2O4; High = NO2; Room temp = equilibrium is established and N2O4 = NO2
f) If they are in separate sealed tubes, the system is closed so they would eventually reach equilibrium
and achieve the medium red-brown color if left alone at room temp.
Evidence:
NO2 = red/brown;
N2O4 = clear;
N2O4 (g) ↔ 2NO2 (g)at equilibrium = medium red-brown