Question

1. The conversion of P to Q is an important cellular reaction. The reaction is reversible. product The equilibrium constant (K'ea) is: Pleq ct ant Suppose [P]pr-3 M and [Q]pr-10 M. LP 3.33 a. In words, what does a K'eq of 0.2 mean? t means that at equilibrium Peq has 5 tmes the concentration of [G]eq b. Given these prevailing conditions, what is the sign of AG' for the forward reaction (P-Q)? Show your work or briefly justify your response. [a7 LPpr Energy spontaneous exergo FND c. Given these prevailing conditions, what is the sign of AG' for the reverse reaction (QP)? Show your work or briefly justify your response. 2 AGEnerndeased 3 Rvs Would the forward or reverse reaction occur spontaneously under these prevailing concentrations of P and Q? Show your work or briefly justify your response. ке d. e. What is the sign of AG°" for the forward reaction (P->Q)? Show your work or briefly justify your response. 1standard state

Answer 1

a) Keq is the ratio concentrations of product to the concentration of reactant so here keq is 0.2 what you have written is correct that at an equilibrium concentration of P is 5 times to that of Q.

b) we have to find out $\Delta$G`

$\Delta$G`= $\Delta$G<sup>o</sup>`+2.303RTlog[Q]/[P]

first finding $\Delta$G^o`

$\Delta$G^o`=-2.303RTlog Keq

=-2.303*8.314*298*log 0.2 (R is universal gas constant 8.314 J/mol.K and temperature T is 298K)

=3988.2 J

Now $\Delta$G`= 3988.2+ 2.303*8.314*298*log3.33

= 3988.2+2980

=6969.18J

here value we got is positive so sign of  $\Delta$G` is positive for forward reaction.

Now use the above equations for standard free energy change and free energy change further

c) for backward reaction

keq= [p]/[Q]

= 5

K= 1/3.33

=0.3

now

$\Delta$G^o`= -2.303*8.314*298*log5

=-3988.2J

$\Delta$G`= -3988.2+2.303*8.314*298*log.3

=-3988.2-2983.4

=-6971.66J

here the sign of free energy change is negative.

d) as calculated in b) we got the free energy change of forward reaction as positive which means energy is required for the reaction to occur so the reaction is not spontaneous