Page 5 of 8 VI) (10 points) For a certain chemical emperature (in °C) for the re
ID: 696993 • Letter: P
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Page 5 of 8 VI) (10 points) For a certain chemical emperature (in °C) for the reaction to be spontaneous? Assuming that the values ofaH and as are independent f temperature, calculate the equilibrium constant (K) at 50°C reaction, -14.9 kJmo, as-50 JKrmol, what is the lowest (VIID (10 points) The radioactive potassium-40 isotope decays to argon-40 with a half-life of 1.2 x 10 yT.(a) Write a balanced equation for the reaction. (b) Starting with 3.0x 10' kg of potassium-40, calculate the of argon-40 produced after 2.0 x 10 yr.Explanation / Answer
DG0 = DH0-TDS0
for the process to be spontaneous,DG0 = at least 0 or -ve.
0 = 14.9 - (T*50*10^-3)
T = 298 k
if T = 50 c = 323.15 k
DG0 = 14.9 - (323.15*50*10^-3)
= -1.26 kj/mol
DG0 = -RTlnk
-1.26*10^3 = -8.314*323.15lnk
k = equilibrium constant = 1.6
7) 40-K19 ----> 40-Ar18 + 0-e+1(positron)
first order kinetics,
k = 0.693/t1/2
= 0.693/(1.2*10^9)
= 5.775*10^-10 y-1
k = (1/t)ln(a/x)
a = initial amount = 3*10^3 kg
x = amount of isomer 40-k after t, time = x
(5.775*10^-10) = (1/(2*10^9))ln(3*10^3/x)
x = 945.2 kg
amount of 40-K decayed = 3000 - 945.2 = 2054.8 kg
amount of 40-Ar formed = 2054.8 kg
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