Question 26 of 35 Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled 0.0560 mol O2 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 N:O(g) + 3 O2(g) = 4 NO:(g) NEXT 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 NO:(g) + Initial (M) Change (M) Equilibrium (M) 5 RESET 0.0560 0.200 -1.00 -0.0560 0.0200 -0.0200 0.0050 0.0150 -0.0150 0.210 0.190 0.0100 -0.0100 -0.0050 0.0410 0.0710 0.0360 0.180

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Question 26 of 35
Consider the equilibrium system described by the chemical reaction below. A 1.00 L
reaction vessel was filled 0.0560 mol O2 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 N2O(g) + 3 O2(g) =4 NO2(g)
1
NEXT
へ
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)
RESET
0.0560
0.200
-1.00
-0.0560
0.0200
-0.0200
0.0150
-0.0150
0.210
0.190
0.0100
-0.0100
-0.0050
0.0050
0.0410
0.0710
0.0360
0.180
19
Transcribed Image Text:Question 26 of 35 Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled 0.0560 mol O2 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 N2O(g) + 3 O2(g) =4 NO2(g) 1 NEXT へ 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) RESET 0.0560 0.200 -1.00 -0.0560 0.0200 -0.0200 0.0150 -0.0150 0.210 0.190 0.0100 -0.0100 -0.0050 0.0050 0.0410 0.0710 0.0360 0.180 19
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