Question

Rate of decomposition of a substance increases by a factor of 2.25 for a 1.5 times increase in concentration at constant temperature. Calculate the order of the reaction.

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Answer

To determine the order of the reaction given that the rate of decomposition increases by a factor of 2.25 when the concentration is increased by 1.5 times, we can follow these steps:

  1. Understand the Rate Law: The rate law for a reaction can be expressed as: $$ \text{Rate} = k [A]^n $$ where ( k ) is the rate constant, ( [A] ) is the concentration of the reactant, and ( n ) is the order of the reaction.

  2. Given Information:

    • The rate increases by a factor of ( 2.25 ).
    • The concentration increases by a factor of ( 1.5 ).
  3. Setup the Rate Law Equations: Let's denote the initial rate as ( \text{Rate}_1 ) when the concentration is ( [A] ). Then, the rate law is: $$ \text{Rate}_1 = k [A]^n $$ When the concentration increases by a factor of 1.5, the new concentration is ( 1.5[A] ), and the new rate ( \text{Rate}_2 ) is: $$ \text{Rate}_2 = k (1.5[A])^n $$

  4. Relate the Rates: According to the given information, the new rate is 2.25 times the initial rate: $$ \text{Rate}_2 = 2.25 \cdot \text{Rate}_1 $$ Substitute the rate expressions: $$ k (1.5[A])^n = 2.25 \cdot k [A]^n $$

  5. Simplify the Equation: Divide both sides by ( k [A]^n ): $$ (1.5)^n = 2.25 $$

  6. Solve for ( n ): Take the logarithm of both sides to solve for ( n ): $$ n \ln(1.5) = \ln(2.25) $$ Using the values ( \ln(1.5) \approx 0.405 ) and ( \ln(2.25) \approx 0.812 ): $$ n \cdot 0.405 = 0.812 $$ Hence, $$ n = \frac{0.812}{0.405} \approx 2 $$

Conclusion:

The order of the reaction is ( \mathbf{2} ).

[ \boxed{2} ]


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