For the reaction: malate rightarrow fumarate + H_2O Delta G degree = + 3.1 kJ at

Question

For the reaction: malate rightarrow fumarate + H_2O Delta G degree = + 3.1 kJ at 25 degree C If the initial concentrations are: [malate] = 6.2 times 10^-1 M and [fumarate] = 1.7 times 10 M, will the reaction proceed as written? For the reaction: creative phosphate + H_2O rightarrow creatine + P_i Delta G degree = -37.6 kJ at 310 K will the above reaction proceed forwards or backwards when the initial concentrations are: [creative phosphate] 1.2 times 10^-7 M [creative] = 7.8 times 10^-1 M [P_i] = 9.2 times 10^-1 M The enzyme L-glutamate-pyruvate aminotransferase catalyzes the following: L-glutamate + pyruvate rightarrow alpha-ketoglutarate + L-alanine -260.3 kJ At 300 K. predict whether this reaction will proceed spontaneously to the right or left if the starting concentrations are: [L-glutamate] = 3.0 times 10^-5 M [alpha-ketoglutarate] = 1.6 times 10^-2 M [L-alanine] = 6.25 times 10^-3 M [pyruvate] = 3.3 times 10^-4 M

Explanation / Answer

1)
G = 3.1 KJ = 3100 J
T = 25 oC = 298 K
use:
G = -R*T*ln Kc
3100 = -8.314*298*ln Kc
Kc = 0.286

Qc = [fumarate]/[malate]
= (1.7*10^-3)/(6.2*10^-1)
= 2.74*10^-3

since QC < KC, equilibrium will shift to right
Yes reaction will proceed as written

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