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monitta
4 years ago
13

1) The equilibrium constant Kc of the reaction

Chemistry
1 answer:
vfiekz [6]4 years ago
7 0

Answer:

a) 0.30

Explanation:

Kc = [C] [D] / ([A] [B])

Set up an ICE table:

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\A&0.4&-x&0.4-x\\B&0.4&-x&0.4-x\\C&0&+x&x\\D&0&+x&x\end{array}\right]

Plug in values and solve:

9 = x² / (0.4 − x)²

3 = x / (0.4 − x)

1.2 − 3x = x

1.2 = 4x

x = 0.3

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Answer:

t = 5.7634 s

Explanation:

  • A → Pdts
  • - rA = K (CA)∧α = - δCA/δt

∴ T = 400°C

∴ α = 1 ....first-order

∴ CAo = 0.950 M

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⇒ - δCA/δt = K*CA

⇒ - ∫δCA/CA = K*∫δt

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⇒ t = Ln(CAo/CA) / K

⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)

⇒ t = 1.1527 / 0.200 s-1

⇒ t = 5.7634 s

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