Consider the reaction X + 3Y doubleheadarrow 2N + 4Z. A solution was prepared by

Question

Consider the reaction X + 3Y doubleheadarrow 2N + 4Z. A solution was prepared by mixing 200.0mL of 3.56 times 10^-3 M X and 100.0mL 5.78 times 10^-3 Y. After the system had come to equilibrium, the absorbance, A, of the solution was measured at 600 nm where only Z absorbs light. The value of A was found to be 0.733. The following data were also collected for known solutions of pure Z at 600 nm. Calculate the equilibrium concentrations of X, Y, Z and N in the solution prepared in the problem above. Calculate K for the reaction.

Explanation / Answer

for finding the concentration of Z from absorbance data, a plot of concentration vs absorbance will have to be generated.

from the plot A= 488.95* concentration +0.0009

at A for Z = 0.733, 0.733= 488.95* concentration +0.0009

Concentration of Z = (0.733-0.0009)/488.95 = 0.001497M

the reaction is X+2Y ----->2N +4Z

moles of X in the solution = Molarity* Volume(L)= 3.56*10-3*200/10000 =0.000712

moles of Y= 5.78*10-3*100/1000 = 5.78*10-4 moles

Volume after mixing = 100+200= 300ml =0.3

Concentrations : X= 0.000712/0.3= 0.0024, Y= 5.78*10-4/0.3=1.93*10-3M

let M= drop in concentration of X to reach equilibrium

at Equilibrium , X=0.0024-M, Y= 1.93*10-3-2M, Z=4M and, N=2M

given Z=0.001497, 4M= 0.001497, M= 0.001497/4 =0.000374, N=2*0.000374= 0.000748, X= 0.0024-0.000374 = 0.00206, Y= 1.93/1000-0.000374= 0.00155M

K= [N]2 [ Z]4/ [X][Y]2 = (0.000748)2 * (0.001497)4/ (0.00206* 0.00155*0.00155)= 5.7*10-10

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