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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