Question

In: Chemistry

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M....

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.

N2(g)+O2(g)<----->2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

____________M

Expert Solution

the first thing to do is to calculate the equilibrium constant of concentration Kc so:

N2(g)+O2(g)<----->2NO(g).

the equilibrium constant is the relationship between products and reactants so:

Kc = [NO]² / [N₂] [O₂]

calculate Kc with the given numbers:

Kc = [0.5]² / [0.2] [0.2] = 6.25

now analyze the equilibrium with the concentration of NO = 0.8 M, according to lecheteliers principle the equilibrium will shift to the left because you have more of product than the original case

N2(g) + O2(g)<-----> 2NO(g).

I........0.2...............................0.2..........................0.8

C.....+x................................+x............................-x

E......0.2+x...........................0.2+x....................0.8-2x

the equilibrium constant is expressed as:

(0.8 - 2x)² / (0.2+x)(0.2+x) = 6.25

solve this equation to get the value of x, i am using a spreadsheet so x is 0.0666 M

the concentration of NO is

0.8 - 2 * 0.0666 = 0.666 M

*hope it helps =)

queen_honey_blossom answered 2 years ago

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M[N2]=[O2]=0.200 M and [NO]=0.400...

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M[N2]=[O2]=0.200 M and [NO]=0.400 M.[NO]=0.400 M. N2(g)+O2(g)−⇀↽−2NO(g)N2(g)+O2(g)↽−−⇀2NO(g) If more NONO is added, bringing its concentration to 0.700 M,0.700 M, what will the final concentration of NONO be after equilibrium is re‑established? [NO]final=

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M....

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M. N2(g) + O2 (g) <----> 2NO(g) If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established? _____M

PART A.Identify the proper form of the equilibrium-constant expression for the equation N2(g)+O2(g)⇌2NO(g) 1.K=[NO][N2][O2] 2.K=[NO]2[N2][O2] 3.K=[N2][O2][NO]2...

PART A.Identify the proper form of the equilibrium-constant expression for the equation N2(g)+O2(g)⇌2NO(g) 1.K=[NO][N2][O2] 2.K=[NO]2[N2][O2] 3.K=[N2][O2][NO]2 4.K=2[NO][N2][O2] PART B. The equilibrium-constant of the reaction NO2(g)+NO3(g)⇌N2O5(g) is K=2.1×10−20. What can be said about this reaction? 1.At equilibrium the concentration of products and reactants is about the same. 2.At equilibrium the concentration of products is much greater than the concentration of reactants. 3.At equilibrium the concentration of reactants is much greater than that of products. 4.There are no reactants left over once...

Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH...

Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH a) initially b) after addition of 5.00 mL of 0.500 M HCl c) at the half-neutralization point (halfway to equilvalence point) d) after the addition of 10.00 mL of 0.500 M HCl e) at the equivalence point

12. Consider the titration of 0.100 L of 0.200 M ascorbic acid with 0.500 M NaOH....

12. Consider the titration of 0.100 L of 0.200 M ascorbic acid with 0.500 M NaOH. What is the pH at the endpoint of the titration? Enter your answer to two decimal places; for this question we can assume the 5% assumption is valid.

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in...

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.09-L rigid container are [H2] = 0.0496 M, [F2] = 0.0144 M, and [HF] = 0.478 M. H2(g) + F2(g) < == > 2 HF(g) If 0.179 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

If the initial concentration of N2 is 1.1 M and O2 is 1.4 M, what would...

If the initial concentration of N2 is 1.1 M and O2 is 1.4 M, what would the concentration of NO be when equilibrium is established? At high temperatures, nitrogen and oxygen can combine to form nitric oxide, N2(g) + O2(g) ⇌ 2NO(g) Kc = 3.3 x 10-8.

Part A: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5 ) is titrated with 0.500 M...

Part A: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5 ) is titrated with 0.500 M HNO3 . Calculate the pH after the addition of 28.0 mL of HNO3 . Part B: A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5 ) is titrated with 0.40 M NaOH . Calculate the pH after the addition of 33.0 mL of NaOH . Please Explain Steps

Consider the equilibrium N2(g) + O2(g) ⇄ 2 NO(g) At 2300 K the equilibrium constant K...

Consider the equilibrium N2(g) + O2(g) ⇄ 2 NO(g) At 2300 K the equilibrium constant K = 1.7 × 10-3. Suppose that 0.0150 mol NO(g), 0.250 mol N2(g), and 0.250 mol O2(g) are placed into a 10.0-L flask and heated to 2300 K. The system is not at equilibrium. Determine the direction the reaction must proceed to reach equilibrium and the final equilibrium concentrations of each species. to the right to the left [N2] =____ mol/L [O2] = ____mol/L [NO]...

Part A A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3....

Part A A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of HNO3. Part B A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH.