Help needed...please

[samiira]

I think i am screwed for Chem.. i've forgotten so much already.. ive got methods on friday and i am sitting doing chem

anyways.. this question was in Chemology 08

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

which conditions would favour a high yield of methanol?


    Temperature         Pressure

a)  low                        low

b)  low                       high

c)  high                      low

d) high                     high

i would appreciate it if u provide some explanation aswell

[jasoN-]

+ energy indicates the release of energy (or product); hence exothermic.
Ratios: Reactants(3):Products(1)
Therefore an increase in pressure (High -> Low) and a decrease in temperature (Forward for exothermic) would favour a high yield
So the answer is B?

[samiira]

+ energy indicates the release of energy (or product); hence exothermic.
Ratios: Reactants(3):Products(1)
Therefore an increase in pressure (High -> Low) and a decrease in temperature (Forward for exothermic) would favour a high yield
So the answer is B?

yeh dats right... THANX .. if only the answers gave that sorta explanation , life would b much easier

[samiira]

Another Question...

COnsider the following equilibrium equation :

h2(g) + C2N2 (g) <---> 2HCN(g)    Keq = 1.20

Initially, 0.86 mol of H2, 2.8 mol of C2N2 and 1.6 mol of HCN are placed in a 2.0 L flask.
Which of the following is true?

(a) Trial Keq  > Keq  so the reaction proceeds to left
(b) Trial Keq  < Keq  so the reaction proceeds to left
(c) Trial Keq  < Keq  so the reaction proceeds to right
(d) Trial Keq  > Keq  so the reaction proceeds to right

[jasoN-]

By 'Trial' Keq, they simply mean Kq (Kq is the concentration fraction at any point in time, ie. not necessarily at equilibrium)
Given the amount in mol and the volume, work out the concentrations:
[H2] = 0.43M
[C2N2] = 1.4M
[HCN] = 0.8M

Kq = [HCN]^2 / [H2][C2N2]
    = (0.43)^2 / [(1.4)(0.]
    = 0.165
Given Keq = 1.20, more products need to be produced to establish equilibrium, thus as Kq (or Trial Keq) < Keq, the reaction proceeds to the right
Hence C.

[Shark 774]

By 'Trial' Keq, they simply mean Kq (Kq is the concentration fraction at any point in time, ie. not necessarily at equilibrium)
Given the amount in mol and the volume, work out the concentrations:
[H2] = 0.43M
[C2N2] = 1.4M
[HCN] = 0.8M

Kq = [HCN]^2 / [H2][C2N2]
    = (0.43)^2 / [(1.4)(0.]
    = 0.165
Given Keq = 1.20, more products need to be produced to establish equilibrium, thus as Kq (or Trial Keq) < Keq, the reaction proceeds to the right
Hence C.

Yeah right idea, but you messed up your calculation. You used [H2] as [HCN] and vice versa. But yes C is the right answer, the solutions booklet is wrong. A is NOT correct.