One Question , Use this section to ask your question: 1. A student ran the follo

Question

One Question , Use this section to ask your question: 1. A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) + Br2(g) When she introduced 0.146 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of Br2(g) to be 1.77×10-2 M. Kc = B. at 237 K: When she introduced 0.125 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 9.44×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc=

Explanation / Answer

pso nAt 242 K

2NOBr ---------> 2NO + Br2

Intial conc. 0.146 - -

Equlibrium conc. 0.146-2x. 2x x ( let dissociation of NOBr = x).   

Now given conc. of Br2 at eulibrium = 1.77 * 10-2

So x = 1.77 * 10-2

Now Kc = (2x)2 * x / (0.146-2x)2

Putting value of x we get

Kc = (2 * 1.77 * 10-2)2 * 1.77 * 10-2 / (0.146 - 2 * 1.77 * 10-2)2

On solving this

Kc = 1.813 * 10-3

Now at temperature 237 K

2NOBr ---------> 2NO + Br2

Intial conc. 0.125 - -

Equlibrium conc. 0.125-2y 2y y (let dissociation of NOBr = y)

Now given conc of NOBr at equilibrium = 9.44 * 10-2

So 0.125-2y = 9.44 * 10-2

So y = 0.0153

Now Kc = (2y)2 * y / (0.125 - 2y)2

Putting the value of y

Kc = (2 * 0.0153)2 * 0.0153 / (9.44 * 10-2)2

On solving this

Kc = 1.607 * 10-3

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