a) [NOBr] 0.0720 M, 0.0162, 0. 0.0139 M [Noj M. b) [NOBr] 0.121 M, [NO]- 0.0159

Question

a) [NOBr] 0.0720 M, 0.0162, 0. 0.0139 M [Noj M. b) [NOBr] 0.121 M, [NO]- 0.0159 M, [Br2] c) [NOBr] 0.103 M, [NO] 0.0134 M, [Br2] 0.0181 M d) [NOBr] 0.0472 M, [No]- 0.0121 M, [Brz] 0.0105 M 5) Iodine and bromine react to give iodine monobromide, IBr The equilibrium constant, Kc, for this reaction at 150 is 21Br constant for the 12 x 10 what is the equilibrium following equation 4 IBr (g) 43 2 I (g) 2 Brz (g) 6) The equilibrium constant for the reaction (g) CO (g) 43 CH3OH (g) is 1.6 x 10.2 at a certain temperature. if there are 1.17 x 10-2 moles of H2 and 3.46 x 10 moles of CH3OH at equilibrium in a 5.60 L flask, how many moles of Co are present at equilibrium? 7) Predict the sign on AS for the following processes and explain your answer a) C2H2 (g) 2H2 (g) .c2He (g) 1 Calculate the 4Han for the following reaction luag for HCN H30 and CO2 are 108.9 kJlmol, -241.8 ku/mol and -393.5 J/mol respectively

Explanation / Answer

(5) I2 (g) + Br2 (g) <----> 2IBr (g) Kc = 1.2x102

Kc = [IBr]2 / ( [I2][Br2] )

4 IBr <-----> 2 I2 (g) + 2Br2 (g)

Kc' = ( [I2]2[Br2]2 ) / [IBr]4

= 1 / { [IBr]2 / ( [I2][Br2] )}2

= 1 / Kc' 2

= 1/(1.2x102 )2

= 6.94x10-5

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