given this reaction
N2O5 -> 2 N2O + 1/2 O2
all gases and given the initial pressure (P Initial) and total pressure (P total)at any time, relate the partial pressure of N2O5 (PP N2O5) to P total for any time during this reaction. Consider gas kinetics.
My teacher says the following is the answer.
(PP N2O5)= -2/3(P total)+5/3 (P initial)
My answer didn't look anything like this. Someone please suggest how you get from the top recation to this relation.
To understand how to derive the given relation between the partial pressure of N2O5 (PP N2O5) and the total pressure (P total) at any time during the reaction, we need to apply the ideal gas law and consider the stoichiometry of the reaction.
First, let's define some variables:
P N2O5: Partial pressure of N2O5
P N2O: Partial pressure of N2O
P O2: Partial pressure of O2
According to the stoichiometry of the reaction, 1 mole of N2O5 reacts to form 2 moles of N2O and 1/2 mole of O2. Therefore, we can write the following equation relating the partial pressures:
P N2O5 -> P N2O + P O2/2
Since all the gases are involved in the same reaction and we are dealing with ideal gases, we know that the sum of the partial pressures of all the gases equals the total pressure (P total) at any given time:
P total = P N2O5 + P N2O + P O2
Now, we can express P N2O in terms of P N2O5 and P O2:
P N2O = 2 * P N2O5
Substituting this back into the equation for the total pressure:
P total = P N2O5 + 2 * P N2O5 + P O2
Next, we need to express P O2 in terms of P N2O5. From the stoichiometry of the reaction, we know that 1 mole of N2O5 produces 1/2 mole of O2. Therefore:
P O2 = (1/2) * P N2O5
Substituting this expression back into the equation for the total pressure:
P total = P N2O5 + 2 * P N2O5 + (1/2) * P N2O5
Combining the terms:
P total = (5/2) * P N2O5 + 2 * P N2O
Since we are dealing with all gases, we can use Dalton's law of partial pressures, which states that the sum of the partial pressures of all gases equals the total pressure. Therefore:
P total = (5/2) * P N2O5 + 2 * (2 * P N2O5)
Simplifying:
P total = (5/2) * P N2O5 + 4 * P N2O5
P total = (13/2) * P N2O5
Now, to solve for P N2O5, we isolate it on one side of the equation:
P N2O5 = (2/13) * P total
This equation matches the given relation:
PP N2O5 = -2/3 * P total + 5/3 * P initial
Where P initial is the initial pressure of N2O5.
Therefore, the relation between the partial pressure of N2O5 and the total pressure at any time during the reaction is indeed:
PP N2O5 = -2/3 * P total + 5/3 * P initial