Question

The Ke for the following reaction at 225 °C is 1.7 x 102 3H2 (g)+N2(g)2NH3 (g) If the equilibrium mixture contains 0.12 M H2

express answer to 2 sig figs
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Answer #1

Kc = [NH3]2 / [N2][H2]3

Kc = 1.7 x 102

[H2] = 0.12 m

[N2] = 0.022 M

[NH3] = ?

1.7 x 102 = [NH3]2 / [0.022] [0.12]3

1.7 x 102 = [NH3]2 / [0.000038]

[NH3]2 = 0.00646

[NH3] = 0.080 M

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express answer to 2 sig figs The Ke for the following reaction at 225 °C is 1.7 x 102 3H2 (g)+N2(g)2NH3 (g) If th...
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