can some one help me please Taking away reactant or product, will shift the equi

Question

can some one help me please Taking away reactant or product, will shift the equilibrium to replace it. Treat heat as if it was a product (for endothermic reactions) when increasing or decreasing the temperature. Increasing pressure (by decreasing volume) will shift the equilibrium to the side of the reaction with the least number of moles of gas. [Decreasing pressure will have the opposite affect, the reaction shifts to the side with the most moles of gas.] Consider a simple reaction: A(g) doubleheadarrow B (g) where rate_forward = k_f[A] and rate_reverse = k_r[B] What is the distinction between [A]_o and [A] in the column heading? For Trial M_2, calculate the rate of the forward reaction (A rightarrow B) at the equilibrium and the rate of the reverse reaction (B rightarrow A) at equilibrium. How do these values compare? If kf > kr, what are the relative values of the equilibrium concentration B? Explain. In general, how does the rate of the forward reaction (A rightarrow B) at equilibrium compare to the rate of the reverse reaction (B rightarrow A) at equilibrium?

Explanation / Answer

a) [Ao]=initial concentration of A

[A]= equilibrium concentration of A

b)rate forward=kf[A]=0.20*67=13.4

   rate reverse=kr[B]=0.40*33=13.2

At equilibrium rate forward=rate reverse

c) if kf>kr

Then the equilibrium would shift in the forward direction,and more product concentration (B) would be there.

At equilibrium rate of forward reaction =rate of reverse reaction

                     kf[A]=kr[B]

            or Keq=kf/kr=[B]/[A], where keq is the equilibrium constant

if kf>kr, then [B]>[A] from the above expression.

e)At equilibrium rate of forward reaction =rate of reverse reaction

                  

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