For the reaction I2(g)+Br2(g)--->2IBr(g), Kc=280at 150 degrees C. Suppose that 0

Question

For the reaction
I2(g)+Br2(g)--->2IBr(g), Kc=280at 150 degrees C. Suppose that 0.490 mol IBrin a 2.00-L flask is allowed to reach equilibrium at 150 degrees C. What is the equilibrilium concentration of I2? What is the equilibrilium concentration of Br2? For the reaction
I2(g)+Br2(g)--->2IBr(g), Kc=280at 150 degrees C. Suppose that 0.490 mol IBrin a 2.00-L flask is allowed to reach equilibrium at 150 degrees C. What is the equilibrilium concentration of I2? What is the equilibrilium concentration of Br2?

Explanation / Answer

2IBr(g) ----------> I2(g) + Br2(g)

0.490 0 0 (Initial moles)

0.490-2x x x ( Moles at equilibrium) , x = degree of dissociation

(0.490-2x)/2 x/2 x/2 ( molar concentration at equilibrium)

Now, Kc = ([I2] x [Br2])/ [IBr]2

280 = (x/2 . x/2 )/ { (0.490-2x)/2}2

280 = x2/ (0.490-2x)2

280 = {x/(0.490-2x)}2

16.7332 = x/ (0.490-2x) [ Taking square root]

x = 16.7332( 0.490 - 2x)

x = 8.2 - 33.46x

x = 8.2/34.46 = 0.23796 M

equilibrium concentration of I2 and Br2 = 0.23796 M

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