31. Carbonyl bromide decomposes to carbon monoxide and bromine. COBr(g) ->CO(g)+

Question

31. Carbonyl bromide decomposes to carbon monoxide and bromine. COBr(g) ->CO(g)+Brg) K, is 0.19 at 73 °C. If an initial concentration of 0.63 M COBr, is allowed to equilibrate, what are the equilibrium concentrations of COBr,CO, and Br? a. [COBr:]- 0.11 M, [CO]-0.26 M, [Bra]-0.26 M b. [COBn]-0.28 M, [CO]-0.35 M, [Bn]-0.35 M c. d, [COBr1-0.30 M, [C01-0.33 M, [Bn]-0.33 M [COBn] = 0.37 M, [CO]-0.26 M, [Bn]-0.26 M [COBnJ-063 M, [CO]-0.35 M, [Br)-0.35 M e. 32. For the following reaction, 250dg) + O2(g) 2 SOdg) 2 SO(g) the equilibrium constant, Kp, is 0.758 at 627 °C. What is the equilibrium constant, at 627 °C, for the reaction below? So(g) ->SO(g)+ 1/20(g) a. 0.660 b. 0.871 c. 1.15 d. 1.32 e. 1.74 33. The equilibrium constant (K) for the following reaction is 3.94 x 101 at a given temperature. N-(g)+2 H:0(g) -> 2 NO(g)+2 H(g) What is the equilibrium constant for the reaction below at the same temperature? 3 N:(g) + 6 H:0g) ->6 NO(g)+6 H(g) a. 6.12 x 10- b. 1.31 x 10 c. 3.94 x 103 d. 1.18 x 10 e. 1.58 x 101 Chem 1018 Fall 2016

Explanation / Answer

Ans 31 : e) [COBr2] = 0.63 M , [CO] = 0.35 M , [Br2] = 0.35 M

The equillibrium constant Kc is given as :

Kc = [CO] [Br2] / [COBr2]

Putting all the values we get :

= (0.35) (0.35) / (0.63)

= 0.19

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