Consider the titration curve describing the titration of a strong acid by additi
ID: 1071115 • Letter: C
Question
Consider the titration curve describing the titration of a strong acid by addition oft strong base. Which of the following statements is false? a. The pH does not increase very rapidly at the beginning of the titration b. At the equivalence point, the only species present are water and a neutral salt C. The pH increases rapidly around the equivalence point. d. The pH is lower at the end of the titration than at the beginning. e. The equivalence point is at pH = 7.0. Calculate the activation energy of a reaction if the rate constant is 0.75 s^-1 at 25 degree C and 11.5 s^-1 at 75 degree C. a. 47.1 kJ/mol b. 31.4 kJ/mol c. 681 J/mol d. 20.4 kJ/mol e. 15.8 kJ/mol. A reaction has an activation energy of 40 kJ and an overall energy change of reaction of -100 kJ. In each of the following potential energy diagrams, the horizontal axis is the reaction coordinate and the vertical axis is potential energy in kJ. Which potential energy diagram best describes this reaction? Which of the statements concerning equilibrium is false? a. The equilibrium constant usually is independent of temperature. b. The value of the equilibrium constant for a given reaction mixture is the same regardless of the direction from which equilibrium was attained. c. A system that is disturbed from an equilibrium condition responds in a manner to restore equilibrium. d. Equilibrium in molecular systems is dynamic, with two opposing processes balancing one another. e. A system moves spontaneously toward a state of equilibrium. How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5.500 kJ/mol C_8 H_18? C_8 H_1i + 25/2 O_2 rightarrow 8CO_2 + 9H_2 O a. 7200 kJ b. 55500 kJ c. 4.1 times 10^5 kJ d. 8360 kJ e. 3610 kJExplanation / Answer
11)
Answer: d
pH is higher at the end point as at starting , there is only acid and pH of acid is smaller
12)
K1 = 0.75 s-1
T1 = 25 oC = (25 + 273) K = 298 K
K2 = 11.5 s-1
T2 = 75 oC = (75 + 273) K =348 K
use:
ln (K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln (11.5/0.75) = (Ea/8.314)* (1/298 - 1/348)
2.73 = (Ea/8.314)*(4.821*10^-4)
Ea = 47076 J/mol
=47.1 KJ/mol
Answer: a
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