[Krashy]

Hello guys,

i want to confirm if this exercise is correct the way it is or if it can be simplified. I´m not a native speaker, so if my notation seems weird please let me know.
Here is the link with the exercise (only question 1) https://frp.home.xs4all.nl/CCVX/EN/Tentamen%20CCVS-nov%202015%20EN.pdf
Thank you very much in advance! (Sorry for the long post)

1a)

Kc = [CO2] [H2]
        [CO] [H20]

Kc = [1] [1]   = 1 ≠ 5.10   -> no balance
        [1] [1]


b)

[H2] = 1.39 mol
                 10L

number [H2]         = 1.39 x 6 x 10^23
                          = 8.34 x 10^23
number (H atoms) = 16.68 x 10^23


before (t0) number (H-atoms) = 24 x 10^23

after (t1) number (H-atoms in solution) = 24 x 10^23 - 16.68 x 10^23

-> number (H20) = 7.32 x 10^23
                            2

                        = 3.66 x 10^23 (H20 molecules)

                        = 3.66 mol (H20)
                           6.022

                        = 0.608 mol (H20)

ratio: 1 = CO  = 0.608 mol CO
        1              H20



Kc = Y x 1.39  = 5.1
        0.608 x 0.608


Y = 5.1 x 0.370  = 1.36 mol CO2 ≈ 1.39 mol (because the ratio of H2 and C02 is 1 to 1)
       1.39



c)

This reaction is exothermic to the right, so the forward reaction produces heat. If the equilibrium constant grows, so does the number of reactants and therefore they are producing even more heat/energy, that means the temperature has increased.

[AWK]

1b - use ICE table - this can be done without calculator (addition and subtraction only). You will get in seconds correct answer: 1.39, 1.39, 0.61 and 0.61.

1c
https://en.wiki2.org/wiki/Water-gas_shift_reaction+Newton

[Krashy]

ah all right thank you very much! Is a) okay that way? Seems a bit too easy

[AWK]

I commented only these points with some inaccuracy