D D 4) 80% (a, Sun 5:33 AM Q EE Chrome File Edit View History Bookmarks People W

Question

D D 4) 80% (a, Sun 5:33 AM Q EE Chrome File Edit View History Bookmarks People Window Help O O O Min box x Chem x New Ta http C https:// black board.olemiss.edu 1/courses/Chem 106 Web 1 DAVIS 2014-2015 SUM2/cp5.pdf E App untitled folder € App d If Facebook Twitter W Wikipedia N Yahoo! News Popular Imported From Safari Challenge Problem #5 Consider the following reaction CO(g) 2H2 (g) CH3OH (g) K: 26 at 780°C If you start with the following initial concentrations CO 0.01 M 0.02 M CH3OH OM What are the equilibrium concentrations?

Explanation / Answer

Answer –

We are given, reaction with Kc

CO(g) + 2H2(g) <------> CH3OH(g) ,   Kc = 26

Initial [CO] = 0.01 M, [H2] = 0.02 M , [CH3OH] = 0 M

   We need to put ICE chart for calculating concentration each at equilibrium –

      CO(g) + 2H2(g) <------> CH3OH(g)

I    0.01         0.02                  0

C     -x           -x                   +x

E   0.01-x     0.02-x               +x

Kc = [CH3OH(g)] / [CO(g)] [H2(g)]

26 = x / (0.01-x) (0.02-x)

26 [(0.01-x) (0.02-x)] = x

26(x2-0.03x+0.0002) =x

26x2 -0.78x + 0.0052 = x

26x2 – 0.78x –x + 0.0052 = 0

26x2 -1.78x + 0.0052 = 0

a= 26, b = -1.78 , c = 0.0052

we know the quadratic formula

x = -b+/-b2-4ac / 2a

by placing the value in it and calculate x value

x= 0.00306 and x = 0.0654

so , x = 0.000306

so at equilibrium concentration

[CO(g)] = 0.01-x

              = 0.01-0.00306

               = 0.0069 M

[H2(g)] = 0.02-x

            = 0.02-0.00306

           = 0.017 M

[CH3OH(g)] = x = 0.00306 M

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