Calculate the equilibrium concentrations of all three substances.
Kc=1.7e-3
Q=3.6e-3
[NO]=0.0015 mol/L
[O2]=0.025 mol/L
[N2]=0.025 mol/L
I cannot figure out the concentration for NO...please help! thank you!
With Q>K it means products are too large and reactants too small which means the reaction will move to the product side to achieve equilibrium.
You need to show the reaction to know which Kc we are talking about; i.e., the forward or reverse reaction.
the formula is:
N2(g)+ O2(g) -><- 2NO(g)
I assume the concentrations shown are initial concentrations.
...........N2(g)+ O2(g) <--> 2NO(g)
I.........0.025..0.025.......0.0015
C..........+x......+x........-2x
E.......0.0025+x..0.025+x....0.0015-2x
Kc = (NO)^2/(N2)(O2)
Substitute into Kc expression and solve for x, the evaluate 0.025+x and 0.0015-x.
Post your work if you get stuck.
Note: If you are careful you may not be required to solve a long quadratic equation.
To calculate the equilibrium concentration of NO, we can use the equilibrium constant expression and the given concentrations of the other two substances.
The equilibrium constant expression for the reaction is:
Kc = ([NO]^2 * [O2]) / [N2]
Given values:
Kc = 1.7e-3
Q = 3.6e-3
[O2] = 0.025 mol/L
[N2] = 0.025 mol/L
Now, let's solve for [NO]:
Firstly, substitute the given values into the equilibrium constant expression:
1.7e-3 = ([NO]^2 * 0.025) / 0.025
Next, rearrange the equation to solve for [NO]:
[NO]^2 = (1.7e-3 * 0.025) / 0.025
Simplifying further:
[NO]^2 = 1.7e-3
Now, take the square root to solve for [NO]:
[NO] = √(1.7e-3)
Using a calculator, you can calculate the square root of 1.7e-3. The result will give you the equilibrium concentration of NO.