Use the following reaction for the next 2 questions. For the purposes of these p
ID: 978356 • Letter: U
Question
Use the following reaction for the next 2 questions. For the purposes of these problems deltaGo rxn is related to Kp.
H2O (l) --> H2O (g) delta Go rxn = 8.50 kJ/mol at 25.0oC
Part A: If the initial partial pressure of H2O (g) is 1.45x10-2 atm, what is (non-standard) deltaGrxn? The temperature is 25.0oC.
A) -4.23 kJ/mol
B) -1.99 kJ/mol
C) -10.5 kJ/mol
D) 8.50 kJ/mol
E) 10.5 kJ/mol 28.
Part B: If the initial partial pressure of H2O (g) of 1.45x10-2 atm, how will equilibrium be reached from the current state of the system? The temperature is 25.0oC.
A) The reaction is already at equilibrium.
B) Equilibrium can be reached by the reaction going forward.
C) Equilibrium can be reached by the reaction going backward.
D) It cannot be determined from the information given.
PLEASE EXPLAIN HOW YOU GOT YOUR ANSWERS. I HAVE A TEST OVER THIS IN 2 HOURS
Explanation / Answer
delta G0 = -RT ln Kc
or Kc = e (-delta G0/RT)
Kc = 0.0323
when initially partial pressure = 1.45x10-2 atm
and we have then Kp = [H2O (g)]
= 1.45 X 10-2 ...
so delta G = -RTlnKp
= -RTln(1.45 X 10-2)
= 10.5 Kj/mol
B)
we can see that the equiblirium is is not yet achieved....so reaction will have to go in backward direction so as to attain the same value of equiblirium constant.