The terms q and k describe mass action for reactions at non-equilibrium and equilibrium conditions, respectively. for a forward reaction to being favored in a living cell, which of the following statements must be true?
a. the value of k must be less than that of q.
b. the value of q must be less than that of k.
c. no conclusion can be drawn about q and k for a favorable process.
d. none of the listed statements are true; cells are at equilibrium.

Q should be less than K for the forward reaction to be favoured (option C)

Explanation:

Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

since R and T are always positive :

ΔG<0 if Q<Keq and ΔG>0 if Q>Keq ( thus the reverse reaction is favoured)

therefore Q should be less than K for the forward reaction to be favoured

there are two significant figures.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally