1a. How does a catalyst change the equilibrium of a reaction? |
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1a. How does a catalyst change the equilibrium of a reaction? r A catalyst lowers the...
The best description of how a catalyst makes a reaction proceed at a faster rate is a. A catalyst increases the kinetic energy of the reactant particles. b. A catalyst provides an alternate path for the reaction and effectively lowers the activation energy. c. A catalyst lowers the average energy of the product molecules. d. A catalyst changes the position of equilibrium by shifting the reaction towards the products. e. A catalyst raises height of the energy barrier required for...
A catalyst increases the rate of a reaction by providing a different reaction pathway that lowers only the activation energy. raises only the energy of the products. lowers only the energy of the reactants and products. All of these are affected by the presence of a catalyst.
The activation energy for a reaction is 15 kJ mol-1 at 27° C. A catalyst lowers the activation energy to 10 kJ mol at the same temperature. By what factor is the reaction rate increased? Assume that the reactant concentrations and the pre-exponential factor in the Arrhenius equation are unchanged. O A. 1.5 OB. 1.1 OC. 0.67 O D.7.4 O E. 1.7 x 107 Reset Selection
A catalyst lowers the activation energy for a particular reaction from 75.0 kJ mol−1to 40.0 kJ mol−1. By what factor does the rate constant increase if the catalyst is used at 25.0 °C? Assume that all rate constants obey the Arrhenius equation and that the pre-exponential factors for the uncatalyzed and catalyzed reactions are equal
Suppose that a catalyst lowers the activation barrier of a reaction from 121 kJ/mol to 59 kJ/mol . Part A By what factor would you expect the reaction rate to increase at 25 ∘C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.) Express your answer using two significant figures. <Topic_6_kinetics Exercise 14.81 31 of 32 > A Review | Constants Periodic Table Suppose that a catalyst lowers the activation barrier of a reaction from 121...
Label the following statements about equilibrium as true or false. Adding a catalyst will increase the amount of products. False The concentrations of reactants and products are constant. False The rate of the forward reaction equals the rate of the reverse reaction. True The value of the equilibrium constant will change if the temperature changes False
1a. The reaction of NO and O2 produces NO2. 2 NO(g) + O2(g) → 2 NO2(g) The reaction is second order with respect to NO(g) and first order with respect to O2(g). At a given temperature, the rate constant, k, equals 4.6 x 102 M–2 s–1. When the initial concentrations of NO and O2 are 0.02 M and 0.015 M, respectively, the rate of the reaction is ___ M/s. 1b. The rate constant of a first-order decomposition reaction is 0.014...
1- Identify the changes in the following system at equilibrium when the temperature is suddenly decreased. The reaction is endothermic. PCl5 <> PCl3+Cl2 2- A reaction profile (not to scale!) for the reaction O2+NO2=O3+NO 3- A reaction profile (not to scale!) for the reaction C2H5Br+H-=C2H5OH+Br- l the Relerences to access important values if needed for this question. Identify the changes in the following system at equilibrium when the temperature is suddenly decreased The reaction is endothermic. The value of K...
d listing r. (Individ For a particular first-order reaction, it takes 24 minutes for the concentration of the reactant to decrease to 25% of its initial value, what is the value for rate constant (in S-1) for the reaction? A) 2.0-104 s-1 B)9.6 x 10-4 s-1 C) 1.2 × 10-2 s-1 D)5.8 102s-1 12. In the first order reaction A → products, [A1-0.400 M initially and 0250 M after 15.0 mun, what willl after 175 min? e A) 1.67-10-3 M...
5. How much faster would a reaction be if a catalyst is used that lowers the activation energy from 20.0 kJ/mol to 10.0 kJ/mol? Do the calculation at two temperatures: first at 25.0°C and then at 0.0°C. (20 points) (V) Helpful Stuff Thermodynamics: AG° = AH-TAS Nernst Equation: 6 = 6 - (RT/nF)InQ AG=RTIK At 25°C: 8 = 6 - (0.0591/n)logQ AGRT Ke=e AGⓇ =-nF8° Units/Constants: Volt: 1 V = 1 J/C Faraday: 1 F = 96,485 C/mole Arrhenius Equation:...