The value of Kc for the reaction: N2O4(g) ↔ 2NO2 (g) is 0.21 at 373K. If...

The value of Kc for the reaction:

N₂O₄(g) ↔ 2NO₂ (g)

is 0.21 at 373K. If a reaction vessel at that temperature initially contains 0.030M NO₂ and 0.030M N₂O₄, what are the concentrations of the two gases at equilibrium? First, the reaction quotient (Q) [page 645 of textbook] must be calculated, and then use the I.C.E table [page 648-650 of the textbook] to determine the equilibrium concentrations.

Expert Solution

N₂O₄(g) ↔ 2NO₂ (g)

initial 0.03 0.03

change -x +2x

at equlibrium 0.03-x 0.03+2x

Kc = [NO2]² /[N2O4]

0.21 = (0.03+2x)²/0.03-x

0.0063-0.21x = (0.03)² + 4x² + 0.12

4x² + 0.33x-0.0054 =0

x=0.0139

[N2O4] = 0.03-x

= 0.03-0.0139 = 0.0161M

[NO2] = 0.03+2x

= 0.03+2*0.0139 = 0.0578M

queen_honey_blossom answered 2 years ago

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The value of Kc for the reaction: N2O4(g) ↔ 2NO2 (g) is 0.21 at 373K. If...

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