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When the volume is halved for a system at equilibrium that contains 1.85 M H2, 1.36 M N2, and 2.91e-3 M NH3 with an initial Kc at 9.83e-7, how can you determine the new concentrations for N2, H2, and NH3?
The new concentrations can be determined by using PV = nRT. If V is halved, so P(V/2) = nRT, making P = (2n/V)RT, the concentrations are doubled, making the new concentrations to be 3.7 M H2, 2.72 M N2, and 5.82 x10^-3 M NH3.
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