Using this data,
2 NO(g) + Cl2(g) == 2 NOCl(g) Kc = 3.20 X 10-3
NO2(g) == NO(g) + ½ O2(g) Kc = 3.93
calculate a value for Kc for the reaction,
NOCl (g) + ½ O2 (g) == NO2 (g) + ½ Cl2 (g)
I do not understnad how I am suppose to use the Kc values given if they give you a new reaction??
NOCl+1/2 O2 is the same as
2NO+Cl2+1/2O2>>NO2+1/2 Cl2 if you work out the coefficents. Thus, you have a new K which is the product of the Kc's
So am I suppose to use the Kc values that are given??
yes.
To calculate the value of Kc for the given reaction, you can use the concept of equilibrium constants and the relationship between the equilibrium constants of different reactions.
The given reaction is:
NOCl (g) + ½ O2 (g) == NO2 (g) + ½ Cl2 (g)
First, you need to write the balanced equations for the given reactions:
1) 2 NO(g) + Cl2(g) == 2 NOCl(g)
Kc = 3.20 x 10^(-3)
2) NO2(g) == NO(g) + ½ O2(g)
Kc = 3.93
Now, let's consider the reaction you want to calculate the equilibrium constant for. Notice that it can be obtained by adding the two given equations together. However, the reaction equation involves NOCl, NO2, O2, and Cl2, which are not directly present in the given equations.
To proceed, you can use the principle of multiplying and dividing reactions to get the desired equation. Multiplying the first equation by 2 and the second equation by 2 will allow the common species (NO and NO2) to cancel out, leaving you with the desired equation.
1) 4 NO(g) + 2 Cl2(g) == 4 NOCl(g)
Kc' = (3.20 x 10^(-3))^2 (by multiplying Kc by itself, as the equation has been multiplied)
= 10.24 x 10^(-6)
2) 2 NO2(g) == 2 NO(g) + O2(g)
Kc'' = (3.93)^2 (by multiplying Kc by itself, as the equation has been multiplied)
= 15.40
Now, if you add the above two equations together, the common species (NO and NO2) will cancel out, and the desired equation will be left:
4 NOCl (g) + 2 O2 (g) == 2 NO2 (g) + 2 Cl2 (g)
The equilibrium constant for the above reaction will be the product of the equilibrium constants of the added equations:
Kc (desired) = Kc' x Kc''
= (10.24 x 10^(-6)) x (15.40)
= 1.5776 x 10^(-4)
Therefore, the value of Kc for the reaction is approximately 1.5776 x 10^(-4).