Given: N_2(g) + 3H_2(g) 2NH_3(g) At equilibrium at a certain temperature, the co

Question

Given: N_2(g) + 3H_2(g) 2NH_3(g) At equilibrium at a certain temperature, the concentration of NH_3(g), H_2(g) and N_2(g) are 0.980 M, 1.53 M and 0.510 M, respectively. Calculate the value of K, for this reaction. A. 0.526 B. 0.837 C. 1.26 D. 3.02 E. 1.96 At 600 degree C, the equilibrium constant for the reaction 2HgO(s) 2Hg(l) + O_2(g) is 2.8. Calculate the equilibrium constant for the following reaction at the same temperature 1/2O_2(g) + Hf(l) HgO(s) A. -1.7 B. 0.60 C. 0.36 D. 1.1 E. 1.7 Consider the following reaction Ni(CO)_4(g) Ni(s) + 4CO(g) K_c = 0.0309 A mixture of Ni(CO)_4(g) and CO(g), each with a concentration of 0.800 M, and an excess of Ni(s) were confined in a 1.0-L container at 300 K. Which of the following is true? A. Ni(s) is formed. B. Ni(CO)_4(g) forms until equilibrium is reached. C. some Ni(CO)_4(g) decomposes to CO(g) and Ni(s). D. no change occurs since we are at equilibrium. E. K_c changes to 0.51. Consider the following reaction: 2NOCI(g) 2NO(g) + CI_2(g) If the initial concentration of NOC1(g) is 2.5 M, at equilibrium [CI_2] = 0.60 M. Calculate the equilibrium concentration of NOCI(g). A. 1.3 M B. 1.2 M C. 0.70 M D. 1.9 M E. 2.2 M

Explanation / Answer

1. The reaction is N2 + 3H2 ---------> 2NH3

Kc = [NH3]2/([N2][H2]3)

= [0.980]2/([0.510][1.53]3)

= 0.9604/1.827 = 0.526

Answer is A.

2. K for the reaction is 2.8

For reaction,

2HgO -------> 2Hg + O2

K = [Hg]2[O2]/[HgO]2

For reaction,

1/2 O2 + Hg -------> HgO

K' = [HgO]/([O2]1/2[Hg]) = (1/K)1/2

K' =  (1/K)1/2 = (1/2.8)1/2 = 0.598 = 0.60

Answer is B.

3. As in the reaction conditions, Ni is kept in excess, so the reaction will start moving in opposite direction according to Le-Chatelier principle. So, Ni(CO)4 is formed until the equilibrium is reached.

Answer is B.

4. The reaction is:

2NOCl -----> 2NO + Cl2

we need to prepare ICE table for this.

_______________________[NOCl]____________________[NO]___________________[Cl2]

Initial__________________2.5 M______________________0______________________0

Change_________________-2x_______________________+2x___________________+x

Equilibrium____________2.5-2x_______________________2x___________________x

Equilibrium concentration of Cl2 = x = 0.60 M

At equilibrium, [NOCl] = 2.5-2x = 2.5-2(0.60) = 1.3 M

Answer is A.

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