68) Why do reactions tend to proceed at a faster rate as T increases? Answer: Th
ID: 1044392 • Letter: 6
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68) Why do reactions tend to proceed at a faster rate as T increases? Answer: The number of molecular collisions that possesses sufficient energy to barrier to reaction increases. Section: 57 69) The Arrhenius equation models how the rate constant k A) increases as both Ea and T increase B) increases most when Ea increases and T decreases C) increases most when Ea decreases and T increases D) increases as both Ea and T decrease E) increases as reactant concentrations increase Answer: C Section: 5-8 coordinate diagram is 70) Which of the following correctly describes the reaction whose reaction shown? ? Progress odon- A) endergonic with no transition state B) exergonic with no transition state C) endergonic with a transition state D) exergonic with a transition state energy of product less than energy of reactant E) endergonic with an intermediate Answer: D Section: 5-8 71) Given an activation energy of 15 kcal/mol, use the Arrhenius equation to estimate how much faster the reaction will occur if the temperature is increased from 100 'C to 120 "C. R-1.987 cal/mol K. Answer: The reaction will occur about 2.8 timies faster. Section: 5-8 Use K- e- Ea/RTExplanation / Answer
I believe you need the explanations, since the answers are already provided.
69) The Arrhenius equation is given as
ln K = -Ea/R*1/T + ln A
where Ea = activation energy of the reaction, T is the absolute temperature of the reaction and A = collision factor or pre-exponential factor.
Since A is constant for a particular reaction, we have,
ln k ? –(Ea/T)
A smaller value of Ea and a larger value of T will mean that ln k increases (note that there is a negative sign; hence a smaller Ea and larger T means that the value of the negative term is less and hence, ln k is high). Therefore, (C) is the correct answer.
70) The left hand side of the plot denotes the energy of the reactant(s) while the right hand side denotes the energy of the product(s). The energy profile diagram of the reaction shows that the energy of the reaction has a maxima. This maxima corresponds to the energy of the transition state. The transition state is a high energy state (less stable) and quickly decomposes to yield the product(s). The energy of the reactant(s) is lower than the energy of the product(s), i.e, energy is released during the reaction. Such a reaction is known as an exergonic reaction. Therefore, (D) is the correct answer.
71) We have
K = e(-Ea/RT)
Take logarithm on both sides,
ln K = -Ea/RT
Therefore, if K1 and K2 are the rate constants at temperatures T1 = 100°C = (100 + 273) K = 373 K and T2 = 120°C = (120 + 273) K = 393 K, we have,
ln K1 = -Ea/R*1/(373 K) …..(1)
ln K2 = -Ea/R*1/(393 K) …..(2)
(2) – (1) gives
ln K2 – ln K1 = -Ea/R*[1/(393 K) – 1/(373 K)]
====> ln K2/K1 = -Ea/R*(0.002544 – 0.002681) K-1
====> ln K2/K1 = -Ea/R*(-0.000137 K-1)
====> ln K2/K1 = (15 kcal/mol)/(1.987 cal/mol.K)*(0.000137 K-1)
====> ln K2/K1 = (15 kcal/mol)*(1000 cal)/(1.987 cal/mol.K)*(0.000137 K-1) = 1.0342
====> K2/K1 = exp(1.0342) = 2.8128 ? 2.8
The reaction will occur 2.8 times faster at 120°C than at 100°C (ans).
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