Unit-9 Chemical Equilibrium

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Unit-9 Chemical Equilibrium

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A. grams per litre
B. moles per litre
C. atoms per molecule
D. litres per mole
Answer: Active mass or molar concentration is expressed in moles per litre (mol/dm3), which is used in equilibrium constant expressions.
A. [CO2] only
B. [CaCO3]/[CaO]
C. [CaO]/[CaCO3]
D. [CO2][CaO]/[CaCO3]
Answer: Since CaCO3 and CaO are pure solids, they are omitted, leaving Kc equal to the concentration of CO2 gas only.
A. only the forward reaction occurs
B. only the reverse reaction occurs
C. both forward and reverse reactions take place
D. neither reaction takes place
Answer: The double arrow indicates that both forward and reverse reactions proceed simultaneously until equilibrium is reached.
A. shift the equilibrium towards the products
B. shift the equilibrium towards the reactants
C. have no effect
D. decompose SO3 completely
Answer: The product side has fewer gas moles (3 → 2), so increasing pressure shifts the equilibrium towards SO3.
A. shift towards SO3
B. shift towards SO2 and O2
C. be unaffected
D. increase the value of Kc
Answer: Reducing pressure favours the side with more gas moles (the reactant side), shifting equilibrium towards SO2 and O2.
A. irreversible
B. reversible
C. instantaneous
D. non-reactive
Answer: The Haber reaction N2 + 3H2 ⇌ 2NH3 is reversible, so equilibrium is established under suitable conditions.
A. [NH3]^2 / ([N2][H2]^3)
B. [N2][H2]^3 / [NH3]^2
C. [NH3]^3 / ([N2][H2]^2)
D. [N2][H2] / [NH3]
Answer: Kc = [NH3]^2 divided by [N2] times [H2]^3, using the coefficients as exponents.
A. nitric acid
B. ammonia
C. sulphuric acid
D. hydrochloric acid
Answer: The Haber process manufactures ammonia by reacting nitrogen with hydrogen under high pressure and moderate temperature.
A. shifts forward
B. shifts backward
C. remains unchanged
D. stops
Answer: Decreasing volume increases pressure, and the equilibrium shifts towards the side with fewer gas moles, which is the forward (NH3) side.
A. Pressure change affects the equilibrium position
B. Pressure change has no effect on the equilibrium position
C. Catalyst changes the equilibrium position
D. Increasing temperature shifts equilibrium backward
Answer: The number of gas moles is the same on both sides (2 = 2), so pressure has no effect on the equilibrium position.
A. Concentration
B. Pressure
C. Temperature
D. Catalyst
Answer: A catalyst affects only the rate of reaching equilibrium, not the position of equilibrium itself.
A. increases the yield of ammonia
B. shifts the equilibrium backwards, decreasing yield
C. increases the pressure
D. destroys the catalyst completely
Answer: Since the forward reaction is exothermic, too high a temperature shifts equilibrium towards reactants and lowers ammonia yield.
A. static
B. dynamic
C. instantaneous
D. temporary
Answer: Equilibrium is dynamic because the reactions continue in both directions at equal rates even though there is no net change.
A. the concentration of any reactant or product
B. the temperature
C. the pressure of a gaseous system
D. all of the above
Answer: Concentration, temperature, and pressure changes all disturb equilibrium, which then shifts according to Le Chatelier's principle.
A. increased
B. decreased
C. kept constant
D. doubled and halved
Answer: Raising the temperature shifts an exothermic equilibrium towards the reactants, which lowers the value of Kc.
A. have stopped
B. are occurring at equal rates
C. occur only one at a time
D. occur at different rates
Answer: Dynamic equilibrium means both reactions are continuously occurring but at equal rates, giving constant concentrations.
A. A catalyst shifts equilibrium to the products side
B. A catalyst shifts equilibrium to the reactants side
C. A catalyst only speeds up the attainment of equilibrium
D. A catalyst changes the value of Kc
Answer: A catalyst speeds up both the forward and reverse reactions equally, so it only shortens the time to reach equilibrium.
A. It is static
B. It is dynamic
C. It can occur in an open system
D. It stops the reverse reaction
Answer: Chemical equilibrium is dynamic: forward and reverse reactions continue at equal rates in a closed system.
A. increase
B. decrease
C. stay the same
D. become zero
Answer: Adding a reactant shifts the equilibrium towards the products, increasing the amount of SO3 formed.
A. unchanged
B. the reciprocal of the original Kc
C. double the original Kc
D. half the original Kc
Answer: Reversing a reaction inverts the equilibrium constant, so the new Kc is the reciprocal of the original value.
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