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1. ![]()
U zatvorenom sudu odigrava se reakcija prikazana sledećom jednačinom N2(g) + 3H2(g) 2NH3(g)
Pod određenim uslovima, u ravnoteži se nalazi 0,2 mol/dm3 H2 i 0,8 mol/dm3 NH3. Izračunati ravnotežnu koncentraciju N2, ako konstanta ravnoteže ove reakcije iznosi 800 dm6/mol2.
1) 3,33x10-3 mol/dm3 2) 0,2 mol/dm3 3) 5x10-3 mol/dm3 4) 0,1 mol/dm3
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Ravnoteza
N2(g)+3H2(g)----------------->2NH3(g)
K={NH3}(na kvadrat)/{N2}*{H2}(na treci)
800={0.8}*2/{N2}*{0.2}*3
800*{0.2}*3*{N2}={0.8}*2
{N2}=0.64/800*0.008
{N2}=0.64/6.4=0.1 mol/l