Question

I don't understand why x=0.0033. When I put 0.0033 to the 2400=(x)(x) / (6.8*10^-3 -2x), it doesn't work. I get 0.33 for x. What did I do wrong?

  

5. The reaction 2 NO (g) 2 (g) O2 (g) has a value of K- 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L vessel, calculate the equilibrium concentrations of NO, N2, and O2. Make a chart describing relationships in change. 2 NO (g) 2 (g) 2 (g) 6.8 x 10 Initial Change -2x +X +X 6.8 x 103-2x Equil Plug into equilibrium expression and solve for x. 2400 So x 0.0033 [6.8 x 105-2x] Determine Concentrations. 2 NO (g) 2 (g) 2 (g) 6.8 x 103- 2X Equil 0.0033 0.0033 Substitute 0.0001

Answer 1

Dear yes you are wrong,

Actual answer is X = 0.00339 as per below

2400=(x)(x) / {(6.8*10^^-3 )-2x }²

2400 = X² / ( { (4.6 x 10^-5 ) - 0.0272X + 4X² }

0.1104 - 65.28X + 9600X² = X²

9599X² -65.28X + 0.1104 = 0

X² - 0.0068X + ( 1.15 x 10^-5) = 0

(X - 0.00339)² = 0

(X - 0.00339) = 0

X = 0.00339