RP R+P KD =0.520* at 25degree C where the equilibrium constant Ko i s commonly r

Question

RP R+P KD =0.520* at 25degree C where the equilibrium constant Ko i s commonly referred to as the dissociation constars. Assuming one begins with 0.078CM of RP complex (only): Determine the equilibrium amounts of [RP] and [P] at 25 degree C. Suppose that the value of KD changes to 1.47* when the temperature is raised to 37 degree C. Does the equilibrium of part (a) shift to the right (R): shift to the left (L); or remained unchanged (U). Is this process (reaction) sxothzrmc or endothermic or is AH = zero at equilibrium (put correct answer word in the blank)? Starting from the final equilibrium amounts determined in part (a) at 25 degree C. determine the equilibrium amounts of [RP] and [P] if the temperature is raised to 37 degree C. Hints: quad-atics should use ths positive roots and keep three significant digits. If pressed for time, setup r times n table quadratics for partid credit

Explanation / Answer

(a)

      RP ----------------> R + P

     0.078                      0     0   ------------------> initial

     0.078-x                   x       x -----------------> equilibrium

K = [R][P]/[RP]

0.520 = x^2 / 0.078 -x

x^2 + 0.520 x - 0.04056 = 0

x = 0.069

equilibrium concentrations :

[RP] = 0.078 -x = 0.009 M

[R] = x = 0.069 M

(b) value of Kd changes 1.47

equilibrium shifts to the right

(c) process is endothermic

(d)

same like above at 37oC

1.47 = x^2 / 0.078 -x

x^2 + 1.47 x - 0.1147 = 0

x = 0.074

[RP] = 0.078 -0.074 = 0.0004 M

[R] = x =0.074 M

[P] = x = 0.074 M

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