A student ran the following reaction in the laboratory at 684 K: N2(g) + 3H2(g) 2NH3(g)...

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Chemistry

A student ran the following reaction in the laboratory at 684 K:N2(g) + 3H2(g) 2NH3(g) When she introduced 3.26Ã—10-2 moles of N2(g)and 6.07Ã—10-2 moles of H2(g) into a 1.00 liter container, she foundthe equilibrium concentration of H2(g) to be 5.83Ã—10-2 M. Calculatethe equilibrium constant, Kc, she obtained for this reaction.
Kc = __

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4.2 Ratings (499 Votes)

since volume is 1 L, number of moles will be same as concentration

N2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â Â Â Â 3H2Â Â Â Â <------------------> 2NH3
3.26*10^-2Â Â Â Â Â Â Â Â Â Â Â 6.07*10^-2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â (initial)
3.26*10^-2 -xÂ Â Â Â Â 6.07*10^-2 -3xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 2xÂ Â Â Â Â Â Â (at equilibrium)

given
[H2]= 5.83*10^-2 M
so,
6.07*10^-2 -3xÂ Â Â Â = 5.83*10^-2
x = 8*10^-4 M

Kc= [NH3]^2 / {[N2] [H2]^3}
Â Â Â Â Â Â = (2x)^2 / (3.26*10^-2 - x ) (5.83*10^-2)^3
Â Â Â Â Â Â = (2*8*10^-4 )^2 / ((3.26*10^-2 - 8*10^-4 ) (5.83*10^-2)^3)
Â Â Â Â Â Â = (2.56*10^-6)/ (3.18*10^-2 * 1.982*10^-4)
Â Â Â Â Â = 0.406

Answer: 0.406

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