Question 17 of 25 Submit Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled 0.0560 mol Oz and 0.200 mol N2O and allowed to react at 298 K. At equilibrium, there were 0.0200 mol of NO2 present. Determine the concentrations of all species and then calculate the value of Kc for this reaction. 2 N20(g) + 3 O2(g) =4 NO2(g) 1 2 Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. 2 N20(g) + 3 O2(g) = 4 NO2(g) Initial (M) Change (M)

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Chapter13: Chemical Equilibrium
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Problem 112CP: Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were...
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2:25 1
Question 17 of 25
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Consider the equilibrium system described by
the chemical reaction below. A 1.00 L reaction
vessel was filled 0.0560 mol O, and 0.200 mol
N20 and allowed to react at 298 K. At
equilibrium, there were 0.0200 mol of NO2
present. Determine the concentrations of all
species and then calculate the value of Kc for
this reaction.
2 N20(g) + 3 O2(g) =4 NO2(g)
1
2
Based on the given values, fill in the ICE table
to determine concentrations of all reactants
and products.
2 N20(g) +
3 O2(g)
= 4 NO2(g)
Initial (M)
Change (M)
Equilibrium (M)
5 RESET
0.0560
0.200
-1.00
-0.0560
0.0200
-0.0200
0.0100
-0.0100
-0.0050
0.0050
0.0150
-0.0150
0.210
0.190
0.0410
0.0710
0.0360
0.180
Transcribed Image Text:2:25 1 Question 17 of 25 Submit Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled 0.0560 mol O, and 0.200 mol N20 and allowed to react at 298 K. At equilibrium, there were 0.0200 mol of NO2 present. Determine the concentrations of all species and then calculate the value of Kc for this reaction. 2 N20(g) + 3 O2(g) =4 NO2(g) 1 2 Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. 2 N20(g) + 3 O2(g) = 4 NO2(g) Initial (M) Change (M) Equilibrium (M) 5 RESET 0.0560 0.200 -1.00 -0.0560 0.0200 -0.0200 0.0100 -0.0100 -0.0050 0.0050 0.0150 -0.0150 0.210 0.190 0.0410 0.0710 0.0360 0.180
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