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=

Answer 1

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 K_c = (2x)² * x / (0.146-2x)²

Putting value of x we get

K_c = (2 * 1.77 * 10^-2)² * 1.77 * 10^-2 / (0.146 - 2 * 1.77 * 10^-2)²

On solving this

K_c = 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 K_c = (2y)² * y / (0.125 - 2y)²

Putting the value of y

K_c = (2 * 0.0153)² * 0.0153 / (9.44 * 10^-2)²

On solving this

K_c = 1.607 * 10^-3