Question

QUESTION 6 The reaction 21Br(g) 리2(g) + Br2(g) Has an equilibrium constant K 0.012 If the initial concentration of IBr is 0.026, what is the equilibrium concentration of 12(g)? Enter your answer with 2 significant figures. QUESTION 7 For the reaction Initially the concentration of A was 0.2M. After reaching equilibrium, it was found that the concentration of A had decreased by 0.09. What is the value of the equilibrium constant K for this reaction? Enter your answer with 2 significant figures.

Answer 1

Question6

Answer

[I2] = 0.0023

Explanation

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

K = [I2] [Br2] / [IBr]² = 0.012

at equilibrium

[I2] = x

[Br2] = x

[IBr] = 0.026 - 2x

Therefore,

x²/(0.026 - 2x)² = 0.012

x/(0.026 - 2x) = 0.10954

x + 0. 21908x = 0.002848

1.21908x = 0.002848

x = 0.002336

[ I2] = 0.0023

Question7

2A <- - - - - - > 3B

K = [A]²/[B]³

   ^{at equlibrium}

   ^{[A] = 0.2 - 2x}

   ^{[B] = 3x}

^{At equillibrium [A] = 0.2 - 0.09 = 0.11}

   ^{0.2 - 2x = 0.11}

   ^{- 0.2x = - 0.09}

   ^{x = 0.45}

^{[B] = 3x = 3*0.45 = 1.35}

K = (1.35)³/(0.11)²

= 203