1.) For the following reaction, the equilibrium constant Kc is 2.0 at a certain

ID: 519835 • Letter: 1

Question

1.) For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. If the concentration Of both products is 0.10 M at equilibrium, what is the concentration of the starting material, NOBr? 2NOBr(g) <---> 2NO(g) + Br2
*i know the answer is 2.2x10^-2, I'm just really confused on how to get it. Please help with a step by step process :) this would really help me to understand how to do more of these questions. :) 1.) For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. If the concentration Of both products is 0.10 M at equilibrium, what is the concentration of the starting material, NOBr? 2NOBr(g) <---> 2NO(g) + Br2
*i know the answer is 2.2x10^-2, I'm just really confused on how to get it. Please help with a step by step process :) this would really help me to understand how to do more of these questions. :) 1.) For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. If the concentration Of both products is 0.10 M at equilibrium, what is the concentration of the starting material, NOBr? 2NOBr(g) <---> 2NO(g) + Br2
*i know the answer is 2.2x10^-2, I'm just really confused on how to get it. Please help with a step by step process :) this would really help me to understand how to do more of these questions. :)

Explanation / Answer

The reaction is given as

2 NOBr (g) <=====> 2 NO (g) + Br2 (g)

Kc = [NO]2[Br2]/[NOBr]2

At equilibrium, [NO] = [Br2] = 0.10 M and [NOBr] = (x – 2*0.1) M = (x – 0.2) M (x is the initial concentration of NOBr and the factor 2 is introduced because 2 moles of NOBr decomposes into 3 moles of products as shown above).

Plug in values and write

Kc = (0.1)2(0.1)/(x – 0.2)2

===> 2.0 = (0.001)/(x – 0.2)2

===> (x – 0.2)2 = 0.001/2 = 0.0005

===> x – 0.2 = 0.02236 (take only the positive root)

===> x = 0.2 + 0.02236 = 0.22236 0.22

The initial concentration of NOBr is 0.22 M (ans).

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