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Consider the following reaction:2NOBr(g) 2NO(g) + Br2(g)If 0.412 moles of NOBr(g), 0.678 moles of NO, and 0.224 moles of Br2 are at equilibrium in a 10.3 L container at 516 K, the value of the equilibrium constant, Kc, is:

Answer :

Answer: [tex]K_c=0.0587M[/tex]

Explanation: The given chemical reaction is:

[tex]2NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)[/tex]

Equilibrium constant (Kc) in general is written as:

[tex]K_c=\frac{[products]}{[reactants]}[/tex]

Note:- Coefficients are written as their powers

So, the Kc expression for the above reaction will be:

[tex]K_c=\frac{[NO]^2[Br_2]}{[NOBr]^2}[/tex]

Equilibrium moles are given for all of them. Let's divide the moles by given liters to get the concentrations.

[tex][NOBr]=\frac{0.412mol}{10.3L}[/tex]   = 0.040 M

[tex][NO]=\frac{0.678mol}{10.3L}[/tex]   = 0.0658 M

[tex][Br_2]=\frac{0.224mol}{10.3L}[/tex]   = 0.0217 M

Plug in the values in the equilibrium expression to calculate Kc.

[tex]K_c=\frac{(0.0658M)^2(0.0217M)}{(0.040M)^2}[/tex]

[tex]K_c=0.0587M[/tex]

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