Unit-9 Chemical Equilibrium
Explore subjects and practice MCQs
19
Topics
1,900
Total MCQs
Unit-9 Chemical Equilibrium
Answer: The reverse reaction changes the products back into the reactants, and its rate becomes equal to the forward rate at equilibrium.
Answer: The coefficient of HI is 2, so Kc = [HI]^2 divided by the product of [H2] and [I2].
Answer: The word dynamic means both the forward and reverse reactions keep occurring, but at equal rates, so there is no net change.
Answer: Removing a product shifts the equilibrium towards the product side so that more ammonia is produced to replace it.
Answer: The catalyst does not change the equilibrium position but speeds up both reactions so equilibrium is attained faster.
Answer: Finely divided iron (often with promoters) is used as the catalyst in the Haber process to speed up ammonia formation.
Answer: The law of mass action states that the reaction rate is proportional to the product of the active masses (molar concentrations) of the reactants.
Answer: A catalyst provides an alternative path with lower activation energy, making the reaction proceed faster.
Answer: Kc depends on temperature and on the reaction itself, but not on the initial concentrations from which equilibrium is reached.
Answer: Since the catalyst lowers the activation energy for both directions equally, the equilibrium position stays the same.
Answer: At equilibrium both reactants and products are present together, since the forward and reverse reactions balance each other.
Answer: Active mass means the molar concentration of a species, expressed in moles per litre, as used in the law of mass action.
Answer: Kc is written using the molar concentrations of reactants and products raised to their coefficients in the balanced equation.
Answer: Increasing the concentration of the reactant O2 shifts the equilibrium towards the forward direction, producing more SO3.
Answer: Kc changes only with temperature; it is unaffected by changes in concentration or pressure of the system.
Answer: At equilibrium the forward and reverse rates are equal, so concentrations remain constant and the net change is zero.
Answer: Since the forward reaction is exothermic, a low temperature favours SO3, but a moderate temperature is used as a compromise for a reasonable rate.
Answer: Chemical equilibrium is the state where forward and reverse reaction rates are equal and concentrations remain constant.
Answer: The value of Kc indicates how far a reaction proceeds; a large Kc means products dominate and a small Kc means reactants dominate.
Answer: Both sides of this reaction have the same number of gas moles (2 = 2), so a pressure change does not shift the equilibrium.