The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
The order of reaction is defined as the sum of the powers of concentrations In the rate law.
The rate of second order reaction can be expressed as rate = k [A]2
The reaction X → Y follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k [X]2 ________________________ (1)
Let [X] = a mol L-1 , then equation (1) can be written as:
Rate = k [a]2
= ka2
If the concentration of X is increased to three times, then [X] = 3a mol L-1
Now, the rate equation will be:
Rate = k [3a]2
= 9 (ka2)
Hence, the rate of formation will increase by 9 times.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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The order of reaction is defined as the sum of the powers of concentrations In the rate law.
The rate of second order reaction can be expressed as rate = k [A]2
The reaction X → Y follo...
Step-by-step explanation:
• The order of reaction is defined as the sum of the powers of concentrations In the rate law
•
•
• The rate of second order reaction can be expressed as rate = k [A]2
•
Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.
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