3.00 moles of NO2 was initially placed in a 2.00L reaction vessel maintained at 500C. After equilibrium was established, it was found that 25.0% of the NO2 has dissociated into NO and O2. Find the value of Kc.
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Hi Ush!
This is a equilibrium problem and so you must keep in mind that the equation we need is Kc = concentration of products/ concentration of reactants. Now, we must write out the dissociation reaction so that we can see the stoichiometric ratios between the products and reactants: 2NO2-2NO + O2. See how the ratio between NO2 and NO is 1:1 but NO2 and O2 is 2:1? This will affect out equilibrium expression as we have to raise the concentrations to the power of their coefficients.
It should look like this without any variables plugged in:
Kc = [NO]^2[O2]/ [NO2]^2
Now, we know that 25% of the NO2 dissociated and so there will by 75% of it remaining. So if 25% of NO2 was used up, then 25% of each product was made. This is true for NO but not for O2 since remember that it takes 2 NO2 just to make 1 O2 so really it makes 12.5%. Now since equilibrium measures concentration, we can easily take our 3.00 moles and divide it by 2.00L to get our molarity, which is 1.5M. However, this is the initial value of concentration and so during equilibrium, only 75% of this remained in NO2, 25% was made into NO and 12.5% was made into O2. Thus we now have our full equation:
Kc = [0.25x1.5]^2[0.125x1.5]/ [.75x1.5]^2
Now just solve for Kc and you should get about 0.0208 or 2.08x10^2
I hope this helped and feel free to ask more questions!
This is a equilibrium problem and so you must keep in mind that the equation we need is Kc = concentration of products/ concentration of reactants. Now, we must write out the dissociation reaction so that we can see the stoichiometric ratios between the products and reactants: 2NO2-2NO + O2. See how the ratio between NO2 and NO is 1:1 but NO2 and O2 is 2:1? This will affect out equilibrium expression as we have to raise the concentrations to the power of their coefficients.
It should look like this without any variables plugged in:
Kc = [NO]^2[O2]/ [NO2]^2
Now, we know that 25% of the NO2 dissociated and so there will by 75% of it remaining. So if 25% of NO2 was used up, then 25% of each product was made. This is true for NO but not for O2 since remember that it takes 2 NO2 just to make 1 O2 so really it makes 12.5%. Now since equilibrium measures concentration, we can easily take our 3.00 moles and divide it by 2.00L to get our molarity, which is 1.5M. However, this is the initial value of concentration and so during equilibrium, only 75% of this remained in NO2, 25% was made into NO and 12.5% was made into O2. Thus we now have our full equation:
Kc = [0.25x1.5]^2[0.125x1.5]/ [.75x1.5]^2
Now just solve for Kc and you should get about 0.0208 or 2.08x10^2
I hope this helped and feel free to ask more questions!