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

                             CO +    2H2 <---> CH3OH

initial :                  0.01         0.02              0

At equilibrium :    0.01-x      0.02 -2x          x

Kc = [CH3OH] / {[CO][H2]^2}

26 = x / (0.01-x)(0.02-2x)
26 = x / (0.0002- 0.04x +2x^2)
0.0052 - 1.04x + 52x^2 = x
52x^2 - 2.04x + 0.0052 = 0
solvinf for x we get,
x= 0.036 and x = 0.00274
x can't be 0.036 because that will make equilibrium concentration of CO and H2 negative.
so, x = 0.00274

So equilibrium comcentrations are:
[CO] = 0.01-x = 0.01 - 0.00274 = 0.00726 M
[H2]= 0.02 - 2x = 0.02 - 2*0.00274 = 0.01452 M
[CH3OH] = x = 0.00274 M

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