leachons below. Determine the value of the tnbuin constant H2(g) + Br2(g) 2 HBr(

Question

leachons below. Determine the value of the tnbuin constant H2(g) + Br2(g) 2 HBr(g) 2 HBr(g) H2(g) + Br2(g) Kc = 3.8 x 104 Kc = ? A) 6.4 x 10-4 B) 1.6×103 D) 2.6 x 10-5 E)5.3 x 10-5 1) The equilibrium constant is given for one of the reactions below. Determine the value of the C) 1.9 x 104 missing equilibrium constant. 2HD(g) H2(g) + D2(g) 2 H2(g) + 2 D2(g)-4 HD(g) Kc = 0.28 Kc = ? A) 13 B) 1.9 C)7.8 x 10-2 D)3.6 E) 0.53 What is n for the following equation in relating Kc to Kp? N2(g) + 3 H2(g) 2 NH3(g) A) 4 B) 2 C) 1 D) -2 E)-4 What is n for the following equation in relating Kc to Kp?

Explanation / Answer

19) PCl5 (g) <----------> PCl3 (g) + Cl2(g)

the equilibrium constant

K = [PCl3][Cl2] /[PCl5]

Option D

20)

H2 + Br2 ----------> 2 HBr Kc = 3.8x104

Kc = [HBr]2/[H2][Br2] = 3.8x104

2HBr -----------> H2 + Br2

Kc = [H2][Br2]/[Hbr]2

= 1/Kc of previous

= 1/3.8x104

= 2.63 x10-5

Thus if the balanced equation is reveresed in direction the kc of new equation = 1/Kc of previous equation

option D

21)2HD ----------> H2 + D2 Kc = 0.28

Kc = 0.28 = [H2][D2] /[HD]2

2H2 + 2 D2 ----------> 4HD  

Kc (2) = [HD]4 / [H2]2[D2]2

= 1/Kc2

= 1(0.28)2

=12.755

option B

If the equation is multiplied by a factor of x the new Kc = old Kcx

and if the equation is divided by a factor x , the new Kc = old Kc 1/x

22)N2(g) + 3H2 (g) ---------> 2 NH3 (g)

delta n = number of gas products - number of gas reactants

= 2 - (1+3)

= -2

Option D

23) 2SO2 + O2 ---------> 2SO3

delta n = 2 - (1+2)= -1

option a

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