Unit-9 Chemical Equilibrium

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Unit-9 Chemical Equilibrium

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A. SO3
B. H2SO4
C. H2S
D. Na2SO4
Answer: In the Contact process, SO2 reacts with oxygen to form SO3, which is then used to make sulphuric acid.
A. towards the reactants
B. towards the products
C. to the left only
D. towards the catalyst
Answer: Decreasing a product concentration makes the forward reaction more favourable, so the equilibrium shifts towards the products.
A. the products never re-form reactants
B. products can re-form reactants under suitable conditions
C. only the forward reaction takes place
D. only the reverse reaction takes place
Answer: In reversible reactions the products can change back into reactants, and equilibrium is established when both rates become equal.
A. the forward reaction is exothermic
B. the forward reaction is endothermic
C. ammonia freezes at high temperature
D. the catalyst stops working
Answer: The formation of ammonia is exothermic, so high temperature favours the reverse endothermic decomposition and reduces the yield.
A. increases the stress
B. reduces the stress
C. produces more heat
D. stops the reaction
Answer: Le Chatelier's principle states that the equilibrium shifts in the direction that opposes or reduces the applied stress.
A. [products] / [reactants]
B. [reactants] / [products]
C. [products]^coefficients / [reactants]^coefficients
D. [reactants]^coefficients / [products]^coefficients
Answer: Kc is the ratio of the product of concentrations of products (raised to their coefficients) to that of reactants (raised to their coefficients).
A. shift the equilibrium towards the products
B. shift the equilibrium towards the reactants
C. decrease the value of Kc
D. have no effect at all
Answer: Adding more reactant (H2) shifts the equilibrium towards the forward direction to consume the extra hydrogen.
A. SO3 back into SO2 and O2
B. SO2 back into sulphur and oxygen
C. H2SO4 into water
D. O2 into SO2
Answer: The reverse reaction of the Contact process decomposes SO3 back into SO2 and oxygen.
A. SO3 reacts with the catalyst
B. removing a product shifts equilibrium forward
C. SO3 is a reactant
D. SO3 decomposes easily
Answer: Continuous removal of the product SO3 shifts the equilibrium towards the forward direction, producing more SO3.
A. Concentration
B. Pressure
C. Temperature
D. Catalyst
Answer: Temperature changes both the position of equilibrium and the numerical value of Kc, while other factors only affect the position.
A. reversible reactions
B. the equilibrium constant
C. the catalyst
D. combustion
Answer: By applying the law of mass action to the forward and reverse reactions at equilibrium, Guldberg and Waage derived the equilibrium constant expression.
A. Equilibrium shifts forward producing more NH3
B. Equilibrium shifts backward
C. No change occurs
D. The value of Kc increases
Answer: Adding a reactant shifts the equilibrium towards the products, increasing the formation of ammonia without changing Kc.
A. increases
B. decreases
C. stays the same
D. becomes zero
Answer: A catalyst lowers activation energy and speeds up both directions, so equilibrium is reached more quickly.
A. shifts towards SO3
B. shifts towards SO2 and O2
C. is unaffected
D. stops completely
Answer: Lower temperature favours the exothermic forward direction, shifting the equilibrium towards more SO3 formation.
A. increasing the pressure
B. removing NH3 continuously
C. adding more N2 and H2
D. all of the above
Answer: High pressure, removal of the product, and adding reactants all shift the equilibrium forward and increase ammonia production.
A. the rate constant
B. the equilibrium constant in terms of concentrations
C. the catalyst concentration
D. the activation energy
Answer: Kc stands for the equilibrium constant expressed in terms of molar concentrations of reactants and products.
A. increase the yield of ammonia
B. decrease the yield of ammonia
C. have no effect
D. stop the reaction
Answer: Lowering the temperature favours the exothermic forward direction, so more ammonia is formed.
A. keep changing constantly
B. become fixed and constant
C. become zero
D. double continuously
Answer: In a closed system at equilibrium, the amounts of all species become constant because forward and reverse rates are equal.
A. reactants
B. products
C. catalyst
D. solvent
Answer: When Kc is much greater than 1, the products dominate and the reaction proceeds almost to completion.
A. N2(g) + 3H2(g) ⇌ 2NH3(g)
B. H2(g) + I2(g) ⇌ 2HI(g)
C. N2(g) + O2(g) ⇌ 2NO(g)
D. CaCO3(s) ⇌ CaO(s) + CO2(g)
Answer: In the Haber reaction the product side has fewer gas moles (2 vs 4), so increased pressure favours this reaction most.
2 3 4 5