Need some guidance with b please. Prt-la Chapter 15 Classwork: Equilibrium . For

Question

Need some guidance with b please.

Prt-la Chapter 15 Classwork: Equilibrium . For a reaction that has "gone to completion" or "does not occur", will there be any equilibrium/equilibrium constant? Explain. 2. For the following chemical equilibrium H2(g) +12(g) 2HI (g) K4-49.7 Determine the concentration of H2, I2, and HI at equilibrium assuming two different initial conditions. Check your answer for each case by substituting the calculated equilibrium values into the equilibrium expression and verifying that the calculated value of K matches 49.7. a) H2 and I2 are initially both 1.0 M ra 2 Hint: you should not have to use the quadratic formula here. See if you can find the math shortcut. 10 o (i-x) .65)I- ataa.11.Son 78M .os_ 9.0s -ax b) H2 is initially 5.0 M and l2 is initially 0.0001 M S,00 2 HF CR a DM 0.00o1 (So-x t 5.000

Explanation / Answer

At equilibrium, the rate of forward reaction = rate of back ward reaction.

When the reaction is complete , there are no reactants and hence there is no reverse reaction. So there is no equilibrium constant. Similarly, where is no reaction, there is no back ward reaction and rate of the reaction does not arise. but however, this is hypothetical case and any reaction is reversible in nature if not totally, at least to certain extent.

2. for the reaction H2(g)+ I2(g) <-------->2HI(g)

Kc= Equilibrium constant = [HI]2/ [H2][I2]= 49.7

preparing the ICE table

                                              [H2], M                                             [I2], M                                    [HI], M

initial                                         1                                                       1                                              0

change                                     -x                                                      -x                                              2x

Equilibrium                             1-x                                                      1-x                                              2x

KC= (2x)2/(1-x)2= 49.7, x2/(1-x)2= 49.7/4=12.425

taking square root, x/(1-x) = 3.5

x= 3.5-3.5x

4.5x= 3.5, x= 3.5/4.5=0.78

So at equilibrium, [H2]=[I2]=1-0.78=0.22 and [Hi]= 2*0.78= 1.56M

3. for the case where [H2] =5M and [I2]=0.0001 M initially

Preparing the ICE Table

                                             [H2], M                                [I2],M                               [HI], M

initial                                         5                                       0.0001                                0

change                                    -x                                          -x                                      2x

Equilibrium                              5-x                                      0.0001-x                             2x

KC= 4x2/(5-x)*(0.001-x)= 49.7, when solved using excel, x= 0.000999984

at equilibrium, [H2] =4.999 and [I2] =1.61*10-8, and [HI]= 2*0.00099984=0.00199968M

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