At a certain temperature, Kc = 0.914 for the reaction
NO2 (g) + NO (g) <----> N2O (g) + O2 (g)
Equal amounts of NO and NO2 are to be placed in a 5.00 L container until the N2O concentration at equilibrium is 0.050 M. How many moles of NO and NO2 must be placed in the container?
..........NO2 + NO ==> N2O + O2
initial....x.....x.....0......0
equil................0.05....0.05
Kc = 0.914 = (N2O)(O2)/(NO2)(NO)
(N2O) = x
(NO) = x
(N2O) = 0.05
(O2) = 0.05
Solve for x which is molarity NO2 and NO.
Then moles = M x L to solve for moles.
What is the value of equilibrium constant kc for NO2 + N2O =3NO
Given equilibrium concentrations are NO2=1.25,N2O=1.80 and NO=0.0015
To determine the number of moles of NO and NO2 that must be placed in the container, we need to use the given equilibrium constant (Kc) and the desired concentration of N2O at equilibrium (0.050 M). Let's follow the step-by-step process to solve this problem:
1. Start by writing the balanced chemical equation for the reaction:
NO2 (g) + NO (g) <----> N2O (g) + O2 (g)
2. Use the stoichiometry of the balanced equation to write the expression for the equilibrium constant:
Kc = [N2O][O2] / [NO2][NO]
3. Substitute the given equilibrium constant (Kc = 0.914) into the equation:
0.914 = [N2O][O2] / [NO2][NO]
4. Since we are given that the equilibrium concentration of N2O is 0.050 M, we can substitute this value into the equation:
0.914 = (0.050)[O2] / [NO2][NO]
5. Now, let's assign variables to the unknown quantities provided in the problem. Let's say the initial moles of NO2 and NO are "x" and "x" respectively.
[N2O] = 0.050 M (given)
[O2] is the same as [N2O] since the stoichiometric coefficient of N2O is 1.
Substitute the variables into the equation:
0.914 = (0.050)(0.050) / (x)(x)
6. Simplify the equation:
0.914 = 0.0025 / x^2
7. Rearrange the equation to solve for x:
x^2 = 0.0025 / 0.914
8. Take the square root of both sides to solve for x:
x = √(0.0025 / 0.914)
9. Calculate the value of x:
x ≈ 0.054 moles
10. Since we initially assumed that the moles of NO2 and NO are equal, the moles of NO2 and NO that must be placed in the container are both approximately 0.054 moles.
Therefore, approximately 0.054 moles of NO and NO2 must be placed in the 5.00 L container.