At 500.0K, one mole of gaseous ONCl is placed in a one-liter container. At equilibrium it is 5.3% dissociated according to the equation shown here: 2ONCl <==> 2NO + Cl2. Determine the equilibrium constant
i dont even know how to start. please help and provide explanation
thank you so much!!!
i dont even know how to start. please help and provide explanation
thank you so much!!!
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Sincerely.
In the beginning, you have x mol of ONCl (temperature and volume do not really play a role in here). 5.3% dissociated, which means you will have 0.947X in the equilibrium. If 5.3% dissociates, you will have the same amount in NO and the half amount in Cl2. That means 0.053X NO and 0.0265X NO. You have to add up the total amount so you can express their concentration. 0.947X+0.053X+0.0265X = 1.0265X.
So the concentrations are:
[ONCl] = 0.947X/1.0265X = 0.9226
[NO]= 0.053X/1.0265X = 0.05163
[Cl2]=0.0265X/1.0265X = 0.02582
K=[Cl2][NO]/[ONCl]^2 = 0.001566
p.s.: As you see, I did not really use the temperature and the volume of the gas in my solution. They are not neccesary :)
In the beginning, you have x mol of ONCl (temperature and volume do not really play a role in here). 5.3% dissociated, which means you will have 0.947X in the equilibrium. If 5.3% dissociates, you will have the same amount in NO and the half amount in Cl2. That means 0.053X NO and 0.0265X NO. You have to add up the total amount so you can express their concentration. 0.947X+0.053X+0.0265X = 1.0265X.
So the concentrations are:
[ONCl] = 0.947X/1.0265X = 0.9226
[NO]= 0.053X/1.0265X = 0.05163
[Cl2]=0.0265X/1.0265X = 0.02582
K=[Cl2][NO]/[ONCl]^2 = 0.001566
p.s.: As you see, I did not really use the temperature and the volume of the gas in my solution. They are not neccesary :)