I don\'t understand why x=0.0033. When I put 0.0033 to the 2400=(x)(x) / (6.8*10
ID: 475065 • Letter: I
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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 0.0001Explanation / Answer
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
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