a 2.50-mole quatity of NOCL was initially in a 1.50-L reaction chamber at 400 degrees celcius. after equilibrium was established, it was found that 28.0 percent of the NOCL had dissociated.
2NOCL(g) <---> 2NO(g)+ Cl2(g)
2.5mole/1.5l=1.67M 28/100*167=0.46 0.46 the value of x and 27
mole of NOCL I divide x=0.46/2=0.23 Finaly Concentration (NOCL) =1.67-2(0.23)=1.21 and concentration of (NO) =2(0.23)=0.46 and concentration of Cl2=0.23 or kc=(0.46)^2(0.23)/(1.2)^2=3.402*10^-2
Well, well, well, looks like our NOCL decided to have a party and dissociate! But don't worry, I've got the scoop for you.
To start off, let's calculate how much NOCL is left after this dissociation party. Since 28.0 percent of the NOCL has dissociated, we can say that 72.0 percent is still associated and intact. So, we have:
0.720 * 2.50 moles = 1.80 moles of NOCL remaining
Now, to find out how many moles of NO and Cl2 were produced due to this wild dissociation, we can use the stoichiometry of the reaction. According to the balanced equation, for every 2 moles of NOCL, we get 2 moles of NO and 1 mole of Cl2 (it's all about fair play, you know!).
So, for the 1.80 moles of NOCL we had left, we'd have:
1.80 moles * (2 moles of NO / 2 moles of NOCL) = 1.80 moles of NO
1.80 moles * (1 mole of Cl2 / 2 moles of NOCL) = 0.90 moles of Cl2
So, after the dissociation party, we have 1.80 moles of NO, 0.90 moles of Cl2, and 1.80 moles of NOCL still hanging around in the reaction chamber. Talk about keeping things interesting!
To solve this problem, we first need to calculate the initial and final amounts of each species using the given information and the stoichiometry of the reaction.
1. Calculate the initial amount of NOCL:
The number of moles of NOCL is given as 2.50 mol.
2. Determine the initial amount of each species:
Since the balanced equation tells us that 2 moles of NOCL produces 2 moles of NO and 1 mole of Cl2, the initial amount of NO and Cl2 can be assumed to be zero.
Therefore, the initial amount of NOCL is 2.50 mol, the initial amount of NO is 0 mol, and the initial amount of Cl2 is 0 mol.
3. Calculate the amount of NOCL that dissociated:
It is given that 28.0% of the NOCL has dissociated at equilibrium.
Therefore, the amount of NOCL that has dissociated can be calculated as:
Amount of NOCL dissociated = 28.0% of 2.50 mol = 0.28 * 2.50 mol
4. Calculate the final amount of NOCL:
The final amount of NOCL can be obtained by subtracting the amount of NOCL dissociated from the initial amount of NOCL:
Final amount of NOCL = Initial amount of NOCL - Amount of NOCL dissociated
5. Calculate the final amount of NO and Cl2:
The balanced equation states that 2 moles of NOCL produce 2 moles of NO and 1 mole of Cl2. Since the reaction goes to equilibrium, the amount of NO and Cl2 formed will be the same.
Therefore, the final amount of NO and Cl2 would be half the amount of NOCL dissociated:
Final amount of NO and Cl2 = 0.5 * Amount of NOCL dissociated
Now, you can substitute the values to calculate the final amounts of NOCL, NO, and Cl2.
To determine the equilibrium concentrations of the reactants and products in the given reaction, we need to follow these steps:
Step 1: Calculate the initial concentration of NOCL.
Given: 2.50 moles of NOCL and 1.50 L of reaction chamber volume.
Concentration (C) = moles/volume
Initial concentration of NOCL = 2.50 moles / 1.50 L = 1.67 M (Molar)
Step 2: Determine the change in concentration.
Since 28.0% of NOCL dissociated, we should consider 72.0% of NOCL remains at equilibrium.
Change in concentration of NOCL = 1.67 M x 0.72 = 1.20 M
Change in concentration of NO and Cl2 = 2 x (1.67 M x 0.28) = 0.94 M
(Note: In this reaction, the change in concentration of NO and Cl2 are the same)
Step 3: Calculate the equilibrium concentrations.
Equilibrium concentration of NOCL = Initial concentration - Change in concentration
Equilibrium concentration of NOCL = 1.67 M - 1.20 M = 0.47 M
Equilibrium concentration of NO and Cl2 = Initial concentration + Change in concentration
Equilibrium concentration of NO and Cl2 = 0 + 0.94 M = 0.94 M
So, at equilibrium, the concentration of NOCL would be 0.47 M and the concentrations of NO and Cl2 would be 0.94 M each.