Name: Chem 111 Exam III- Chemieal Equilibrium, Acids and Bases Exam THERE ARE 6

Question

Name: Chem 111 Exam III- Chemieal Equilibrium, Acids and Bases Exam THERE ARE 6 PAGES TO THIS EXAM Signiicant Pucures imust be correet. All set -ups imust be shown or no credit will be given Opoints) 1. If 1.50 x 101 moles of all gaseous substanens were initially placed in a 1,50 L container for the following reaction corld + Hda., cod + H,0@ K=0717 a. Which direction will the reaction procoed to reach equilibrium? Products or reactante? Answer: b. What will be the concentration of all substanees present at equilibrium Answer Page 1

Explanation / Answer

a) For the reaction, CO2+H2 ---> CO + H2O

Equilibrium constant expression is given by K = [CO][H2O]/[CO2][H2]

Given that concentratoin of all gaseous substances = 1.5*10^2 moles/1.5 L = 100 M

Now, Q = 100*100/100*100 = 1

Since Q > Kc, reaction moves towards reactants

b) CO2 + H2 ----> CO + H2O

I-- 100   + 100----100----100

C--+x-------+x----- - x---- -x

E-- 100 + x--------- 100-x

Kc = 100-x*100-x/100+x*100+x = 0.717

solving for x = 8.30 M

At equilibrium, [CO] = 100-8.30 = 91.7 moles

[H2O] = 100-8.30 = 91.7 moles

[CO] = 100+8.30 = 108.3 moles

[H2] = 100+8.30 = 108.3 moles

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