I need to know why my solution for this question does not fit. Please point out it and show the correct answer...

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I need to know why my solution for this question does not
fit.
Please point out it and show the correct answer...

I Need To Know Why My Solution For This Question Does Not Fit Please Point Out It And Show The Correct Answer 1 (505.04 KiB) Viewed 38 times

2NH3 (2) troson ↓ 2 [HH] 9.65m H₂67 NH3 (7 initial 0.683 8,80 3.65 3 Change 즐거 रं मे 3 equilion 0.683 + = 8.80=4 3,65-X -² (3.65-9) 3 ---3 K₁ - 2.37 x 10 = CHH₂ 23 3 (N₂] [H₂] ²³² 3 (0.683 + ²) (2x + P.8) $ ma TS +(8.8) B 2 3 © (²31) + 3 (²6) 5.8 +3 - 2 x (53) 223 92 -X*8.8 + 9 0. {X+3 ·7x (8.8) V 8 3 2.37 × 10¹³ (0.683 + 2) = 73 x.-15 20 × 88 73x-15-1.61 87 1×10²) 2.37×10 7 20 × 88 2 X = -526 39₁ At 720°C, the equilibrium constant K, for the reaction N₂(g) + 3H₂(g) → 2NH3(g) is 2.37 × 10-³. In a certain experiment, the equilibrium concentrations are [N₂] = 0.683 M, [H₂] = 8.80 M, and [NH3] = 1.05 M. Suppose some NH, is added to the mixture so that its concentration is increased to 3.65 M. (a) Use Le Châtelier's principle to predict the shift in direction of the net reaction to reach a new equilibrium. (b) Confirm your predic- tion by calculating the reaction quotient Q, and comparing its value with K. INLICH ₂³ 7680M 8.80M 3 7 (32)² = 212 ³ 29 &N X 2 (3.66)-2xxx 3.65 +-+* 2