can someone help me step by step with the mathmatics involved with this, please?

Question

can someone help me step by step with the mathmatics involved with this, please? I mean like your teaching a highschool kid the algebra rules. thank you ??

1) Consider the reaction between nitrogen trioxide and nitrogen monoxide to generate nitrogen dioxide with all species in the gas phase a) Write a balanced chemical equation for this equilibrium reaction (including states of matter). b) Write the equlibrium expression for Ke for this reaction NOT NO NO) c) A flask is charged with 0.575 M nitrogen dioxide. What are the equilibrium concentrations of each species in the flask? (Kc 87.0-This is NOT a quadratic problem LWO1 0.575 No, kc 97.0 NO3 d) A 425.0 mL flask is charged with 5.00 moles of nitrogen monoxide and 6.00 moles of nitrogen trioxide. What is the equilibrium concentration of each species in the flask? (Ke 87.0- This IS a quadratic problem) MOL . 425 L O.?25L 425 NO, 20-

Explanation / Answer

Part a

Balanced equation:

NO3(g) + NO(g) = 2 NO2(g)

Part b

Equilibrium constant expression of the reaction

Kc = [NO2]2/[NO] [NO3]

Part C

Reaction with ICE TABLE

NO3(g) + NO(g) = 2 NO2(g)

I    0.575

C - x - x +2x

E - x - x 0.575 + 2x

Equilibrium constant expression of the reaction

Kc = [NO2]2/[NO] [NO3]

87 = (0.575 + 2x)2/ (-x) (-x)

87 = (0.575 + 2x)2/ (x2)

Take square root on both sides

9.327 = (0.575 + 2x)/ (x)

9.327x = (0.575 + 2x)

9.327x - 2x = (0.575)

7.327x = (0.575)

x = 0.07847 M

Equilibrium concentration of NO3 = - 0.07847 M

Equilibrium concentration of NO = - 0.07847 M

Equilibrium concentration of NO2 = 0.575 + 2*0.07847

= 0.73194 M

Part d

Initial concentration of NO = 5 mol / 0.425L = 11.765 M

Initial concentration of NO3 = 6 mol / 0.425L = 14.117 M

Reaction with ICE TABLE

NO3(g) + NO(g) = 2 NO2(g)

I 14.117 11.765

C - x - x +2x

E ( 14.117- x) (11.765 - x) 2x

Equilibrium constant expression of the reaction

Kc = [NO2]2/[NO] [NO3]

87 = ( 2x)2/ (14.117-x) (11.765-x)

87(14.117-x) (11.765-x) = 4x2

(1228.179 - 87x) (11.765-x) = 4x2

14449.53 - 1228.179x - 1023.55x + 87x2 - 4x2 = 0

83x2 - 2251.729x + 14449.53 = 0

x = 10.417

Equilibrium concentration of NO3 = 14.117- x

= 14.117- 10.417 = 3.7 M

Equilibrium concentration of NO = 11.765 - x

= 11.765 - 10.417 = 1.348 M

Equilibrium concentration of NO2 = 2x = 2* 10.417

= 20.834 M

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