At equilibrium, the total concentration of products equals the total concentrati

Question

At equilibrium, the total concentration of products equals the total concentrations of reactants. At equilibrium, the rate of the forward reaction is the same as the verse reaction. For the chemical reaction, C_2H_4(g) + H_2O(g) C_2H_5OH(g) Kc = 1.3 times 10^2 The product is the concentration of reactants at equilibrium. Less than Greater than Equal to Unknown because more information is needed to determine What is the state of the system if [I_2(g)] = 1.0M and (I(g)] = 1.0 times 10^-3 M I_2(g) 2 I_(g) Kc = 3.8 times 10^-5 The system is at equilibrium The system is at a steady state The system is not at equilibrium and will produce more reactants The system is not at equilibrium and will produce more products Consider the equilibrium reaction: 2SO_2 + O_2 2SO_3 K_1 2CO + O_2 2CO_2 K_2 What is the equilibrium constant. K, for this reaction? 2SO_2 + 2CO_2 2SO_3 + 2CO K = K_1/K_2 K = K_2/K_1 K = K_1 K_2 K = K_1 + K_2 Which of the following is the correct form of the equilibrium following reaction? A_(aq) + B_(g) rightarrow C_(l) + D_(g) K = [A]/[D]

Explanation / Answer

25. Kc = 1.3*10^2

as Kc is >1,the product is greater than concentration of reactants at equilibrium.

answer: b

26. I2(g) <===> 2I(g)

Qc = [I]^2/[I2]

     = (1*10^-3)^2 /1

   Qc = 1*10^-

Qc < Kc. products are favoured.

answer: d

27.

2so2 + O2 <===> 2so3 k1

2co2 <====> 2CO + o2    1/K2
---------------------------
2SO2 + 2CO2 <===> 2SO3 + 2CO
----------------------------

K = K1*(1/K2)

ANSWER: A

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