Ka = 1.8 x 10-5 = [H30\'][OAc]-x2 1.00-? [HOAc] 2.37 pH--log [H,O]--log (4.2 x 1

Question

Ka = 1.8 x 10-5 = [H30'][OAc]-x2 1.00-? [HOAc] 2.37 pH--log [H,O]--log (4.2 x 10-3) 6) Assume you are at equilibrium with [iso] 1.25 M and [butane] - 0.50 M. Now increase the concentration by 1.50 M butane. Determine the [Iso] and [butane] when the system comes to equilibrium again. Solution Q is less than K, so equilibrium shifts right - away from butane and toward Isobutene. Set up ICE table. Use Moles [nbutane] Isobutene] Initial 0.50+1.50 Change -X Equilibrium 2.00-x 1.25 +x 1.25 +x K 2.50 isobutane]1.25+x 2.00 Round Stic M X 1.07 nm At the new equilibrium position, [Butane] = 0.93 M and [isobutene)-2.32 M. Equilibrium has shifted toward Isobutene.

Explanation / Answer

6)

K = [iso]/[butane]

feeding equilibrium concentrations

K = 1.25/0.50 = 2.5

now when [butane] = 1.50 + 0.50 = 2.00 M

ICE chart

      butane     <===> isobutane

I       2.0                       1.5

C      -x                         +x

E    2.0-x                    1.5+x

So,

2.5 = (1.5+x)/(2.0-x)

5 - 2.5x = 1.5 + x

x = 3.5/3.5 = 1.0 M

So,

equilibrium concentration of [iso] = 1.5 + 1.0 = 2.5 M

equilibrium concentration of [butane] = 2.0 - 1.0 = 1.0 M

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