[wendyfd]

The decomposition of urea, (NH2)2CO, in 0.10 M HCl follows the equation

(NH2)2CO (aq) + 2H+ (aq) +H20 --> 2NH4+ (aq) + CO2 (g)

At 60 degree C, k= 5.84 x 10^-6 min^-1 and at 70 degree C, k=2.25 x 10^-5 min^-1. If this rxn is run at 80 degree celsius starting with a urea concentration of 0.0080 M, how many mins will it take to drop to 0.00060 M ?

My professor wrote out the answer but i am having a hard time understanding certain parts. I will indicate below which parts:

ln(k2/k1) = (-Ea/R) x (1/T2 - 1/T1)
ln(2.25 * 10 ^ -5 min^-1/5.84 * 10^ -6 min ^-1 ) = (-E (subscript)a/8.3134 J mol^-1 K^-1) x (1/343 K - 1/333 K )
1.35 = Ea (8.76 * 10 ^-5 K^-1/ 8.314 J mol ^-1 K^-1)
Ea = 1.28 x 10^5 J/mol

Next he uses the eqn again to find "k2" for the part of the question asking for 80 degree celsius and got an answer of k2 =8.02 * 10^-5 min^-1

By using the answers he got he used the rate law eqn

ln(A) = -kt + ln(A) subscript 0
 
thus

ln (0.0080 M/ 0.0006 M) = (8.02 * 10^-5 min^-1 ) * t
solving for t = 32298

i dont understand how he arranged the formula around to get that expression ? what happened to the negative that is suppose to be in front the "kt"
If ln(A) = 0.0006 M and ln(A) subscript 0 = 0.008M shouldnt it be ln(A)/ln(A) subscript 0 ?

[mjc123]

ln A = -kt + ln A₀
kt = ln A₀ - ln A = ln (A₀/A)