ENVIROMENTAL CHEMISTRY 4. [25] For the completely reversible reaction below, the
ID: 575437 • Letter: E
Question
ENVIROMENTAL CHEMISTRY
4. [25] For the completely reversible reaction below, the first order rate constant for the forward reaction is 500 s-1 and the equilibrium constant is 10. A = B (the equal sign can be interpreted here as a two-sided arrow) a) What is the rate constant for the reverse reaction? b) Determine the concentration of A as a function of time for a system with initial concentrations of [A]0 = 1 mM and [B]0 = 0. Give your answer as an equation for [A] as a function of time. c) How long does it take for [A] to come within 5% of its equilibrium value?
Explanation / Answer
the rate of first order reaction of species is given by -dCA/dt= K*CA
where CA is concentration of A and K is rate constant
for a reversible reaction A<--->B
-dCA/dt= K1CA-K2CB (1), where K1, K2 are rate constants for forward and backward reactions and
K= equilibrium constant = K1/K2
at equilibrium, the rate of forward reaction = rate of backward reaction ( i.e -dCA/dt=0)
hence K1CAe= K2CBe, where CAe and CBe are equilibrium concentrations of A and B.
given K1= 500, K= 10 = K1/K2
K2= K1/10= 500/10= 50
the rate of reaction -dCA/dt= K1CA-K2CB
but CA= CAO*(1-XA), where CA= concentration of A at any time and CAO= initial concentration of A
CB= CBO+CAOXA
-d{CAO*(1-XA)}= 500CAO*(1-XA)- 50(CBO+CAOXA), CBO= initial concentration of B=0
dXA/dt= 500-550XA (1), at equilibrium, XA= XAe, and dXA/dt=0, 500-550XAe=0, XAe= 500/550=0.91
when integrated, the equation becomes
-ln(1-XA/0.91)= 550t, (2)
for A to reach 5% within the equilibrium value, XA= 0.95XAe=0.95*0.91=0.8645
Eq.2 becomes, -ln(1-0.8645/0.91)= 550*t
t= 0.005447 seconds
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