*********** PLEASE DO NOT TRY TO SOLVE THIS UNLESS YOU ARE CERTAIN ABOUT THIS TY
ID: 635825 • Letter: #
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*********** PLEASE DO NOT TRY TO SOLVE THIS UNLESS YOU ARE CERTAIN ABOUT THIS TYPE OF "ICE" QUESTION..... PLEASE HELP AND EXPLAIN IN DETAIL ************ I DON'T WANT JUST THE ANSWERS BUT I WANT TO LEARN ********** THANK YOU IN ADVANCE......
Part A The reaction N2(9) +02(9) 2NO(g) is carried out at a temperature at which Ke-0.045. The reaction mixture starts with only the product [NO-0.0200 M and no reactants Find the equilibrium concentrations of N2 at equilibrium. Express your answer to two significant figures and include the appropriate units. You may want to reference (Pages 693-702) Section 15.8 while completing this problem INa] 0.055 SubmitP incorrect; Try Again: 6 attempts remaining The initial conditions include no reactants in the mixture, so the equilibrium will shift from the product to the reactants. Design an ICE (Initial, Change, Equilibrium) table for the reaction shifting toward the reactants (left) to determine the correct expression for the equilibrium. Part B Find the equilibrium concentrations of O2 at equilibrium. Express your answer to two significant figures and include the appropriate units. O2lValue Units Part C Find the equilibrium concentrations of NO at equilibrium Express your answer to two significant figures and include the appropriate units NOValue Units Submit est AnsExplanation / Answer
N2(g) + O2(g) ---------------> 2NO(g)
I 0 0 0.02
C +x +x -2x
E +x +x 0.02-2x
Kc = [NO]^2/[N2][O2]
0.045 = (0.02-2x)^2/x*x
0.045 = (0.02-2x/x)^2
0.212 = 0.02-2x/x
0.212x = 0.02-2x
x = 0.00904
[N2] = x = 0.00904M
[O2] = x = 0.00904M
[NO] = 0.02-2x = 0.02-2*0.00904 = 0.00192M
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