Question

14. A catalyst a. Participates in the reaction and b. Dec c. Increases the activation e d. Binds to the reactants and releases them sequentially the reaction and changes the product release reases the activation energy required for the reaction 15. In an exothermic reaction, if heat is removed from the reaction a. Equilibrium shifts to the right b. Product formation increases c. Increase the value of K d. All the above 16. Initially, a mixture of 0.100 M No, 0.050 M H2 and 0.100 M H20 wa s allowed to reach equilibrium. There was no N2 present initially. 2 NO(g)+2 H2(g)N2(g) + 2 H2o(g). At equilibrium, the concentration of NO was found to be 0.062 M. Kc of the reaction would be a. 6.54X103 b. 65.4X103 c. 0.654X103 d. 654X103 17. The reaction quotient,Q and equilibrium constant K are related in that a. at equilibrium Q-K b. at equilibrium Q>K c. at equilibrium QcK d. There are never related. 18.[C] [DJ Al [Bjb Keq. In this equation, the exponents represent a. Equilibrium constant b. Phase of the reactant c. Coefficients d. Reaction order 19. The following reaction can reach a heterogenous equilibrium a. 2S02(g)+O2(g) 2SOs(g) b. N2O4 (g)-2NO2(g) c. N2(g)+3H2(g)2NH3(g) d. None of the above

Answer 1

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14b decreases the activation energy required for the reaction.

When the activation energy is lowered ,the energy barrier can be overcome faster and hence the products can be formed faster. Thus catalysts speed up the reaction in this way

15)d all of the above .

The equilibrium will shift to the right so that the conditions are restabilised . Since heat is removed , the reaction moves to the side where more heat is produced to compensate for the less heat ie the product side . Since the reaction moves to the product side ,K value also increases(Le Chatlers principle )

16)c

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* First write the equilibrium expression » Create an ICE chart NO H2 N2 H20 Initial Conc 0.100 0.0500 0 0.100 Change in Conc (M) -2x -2 x Equilibrium 0.062 Conc (M) x + 2 x The change in concentration of the NC (0.062 M - 0.100M) -- 0.038 M. Thus -2x-- 0.038 x 0.019

x = 0.019 Keeping stoichiometry in mind change in concentration of hydrogen-- 0.038 M Change in concentration of nitrogen-+ 0.019 M Change in concentration of water-+ 0.038 M. Complete the chart H2 N2 H20 Initial Conc 0.100 0.0500 0 0.100 Change in - Conc (M) 0.038 0.019 0.038 ЕЧиǐlibrium 0.062 0.012 0.019 0.138 Conc (M)

Initial Conc 0.100 0.0500 0 0.100 Change in - Conc (M) 0.038 0.019 0.038 Equilibrium 0.062 0.012 0.019 0.138 Conc (M) » Substitute values into the equilibrium expression and solve for K. [0.0191T0.13812 0.0621 10.01212 650 or 6.5 x 102 -0.65 × 1013

17) a at equilibrium Q=K

18)c Coefficients .They represent the coefficients of the reactants in the reaction

19)d none of the above

A system at equilibrium that has products and reactants in a single phase is in a homogeneous equilibrium; a system in which reactants, products are in different phases is said to be in a heterogeneous equilibrium.