10 CBSE board-based questions and their answers derived from the paragraph you provided on Rate Law and Rate Expression (Class 12 Chemistry – Chapter: Chemical Kinetics):
1. Define rate law.
Answer:
Rate law is an expression in which the reaction rate is given in terms of the molar concentration of reactants, with each term raised to a power. These powers (called order with respect to a reactant) may or may not be equal to the stoichiometric coefficients of the reacting species in a balanced chemical equation.
2. Write the general rate law expression for the reaction:
aA + bB → cC + dD
Answer:
The general rate law expression is:
Rate = k[{A}]^x.[{B}]^y3. What is the differential rate law? Write its expression.
Answer:
The differential rate law expresses the rate of change of concentration of a reactant or product with time. It is written as:
-dR/dt= k{A}^x{B}^y
4. In the reaction 2NO(g) + O₂(g) → 2NO₂(g), the rate law is given as:
Rate = k[NO]²[O₂].
What is the order of the reaction?
Answer:
Order of the reaction = 2 (for NO) + 1 (for O₂) = 3.
5. How can the order of reaction be determined experimentally?
Answer:
The order of reaction can be determined by measuring the initial rate of reaction while varying the concentrations of reactants and observing the change in rate.
6. Based on the given table, what happens to the rate when [NO] is doubled and [O₂] is kept constant?
Answer:
When [NO] is doubled from 0.30 to 0.60 mol/L and [O₂] is constant at 0.30 mol/L, the rate increases from 0.096 to 0.384 mol L⁻¹ s⁻¹, i.e., by a factor of 4.
This shows that the rate is proportional to .
7. What conclusion can be drawn when the concentration of O₂ is doubled while [NO] is kept constant?
Answer:
When [O₂] is doubled from 0.30 to 0.60 mol/L and [NO] is constant, the rate doubles (from 0.096 to 0.192 mol L⁻¹ s⁻¹).
Thus, the rate is directly proportional to .
8. Define rate constant (k).
Answer:
Rate constant (k) is the proportionality factor in the rate law expression that relates the rate of reaction to the concentrations of reactants. Its value depends on temperature and the nature of the reaction.
9. Why can’t we predict the rate law just by looking at the balanced chemical equation?
Answer:
Because the exponents in the rate law (i.e., the orders of reaction) may not be the same as the stoichiometric coefficients. They must be determined experimentally.
10. If the rate law for a reaction is Rate = k[A]²[B], what is the effect on rate if [A] is tripled and [B] is doubled?
Answer:
Rate ∝ [A]²[B]
New rate ∝ (3A)² × (2B) = 9 × 2 = 18 times the original rate.
No comments:
Post a Comment