1. Calculate the activation energy of a reaction whose rate constant 4.22 * 10^-
ID: 1021992 • Letter: 1
Question
1. Calculate the activation energy of a reaction whose rate constant 4.22 * 10^-4 s^-1 at -25 degrees C and 4.54 * 10^-4 s^-1 at 95 degrees C. Please show work for credit.
2. In the reaction 2 A -> X + 2Y, the initial concentrations of reactants is 0.240 M and there are no initial products. If the equilibrium constant is 1.50 * 10^-8, what are the equilibrium concentrations? Please show work for credit.
3. Calculate the pH of a solution containing 0.685 M HC2H3O2(aq) (Ka = 1.75 * 10^-5). Please show work for credit.
4. Calculate the pH of a buffer containing 0.030 M HNO2 and 0.050 M NaNO2. The Ka of HNO2 is 4.6 * 10^-4. Please show work for credit.
Explanation / Answer
1.
T1= -25 oC = -25+273 K = 248 K
T2= 95 oC = (95+273) K = 368 K
K1 = 4.22*10^-4 S-1
K2 = 4.54*10^-4 S-1
use:
ln(k2/k1)=(Ea/R)(1/T1-1/T2)
ln(4.54/4.22 )=(Ea/8.314)* (1/248 - 1/368)
0.07309 = (Ea/8.314)*1.315*10^-3
Ea= 462 J
Answer: 462 J
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