For the reaction:
2A + B → A2B
the rate = k[A][B]2with k= 2.0 x 10-6mol-2L2s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.
The initial rate of the reaction is
Rate = k [A][B]2
= (2.0 × 10 - 6mol - 2L2s - 1) (0.1 mol L - 1) (0.2 mol L - 1)2
= 8.0 × 10 - 9mol - 2L2s - 1
When [A] is reduced from 0.1 mol L - 1to 0.06 mol - 1, the concentration of A reacted = (0.1 - 0.06) mol L - 1 = 0.04 mol L - 1
Therefore, concentration of B reacted= 1/2 x 0.04 mol L-1 = 0.02 mol L - 1
Then, concentration of B available, [B] = (0.2 - 0.02) mol L - 1
= 0.18 mol L - 1
After [A] is reduced to 0.06 mol L - 1, the rate of the reaction is given by,
Rate = k [A][B]2
= (2.0 × 10 - 6mol - 2L2s - 1) (0.06 mol L - 1) (0.18 mol L - 1)2
= 3.89 mol L - 1s - 1
The rate constant for the decomposition of N2O5 at various temperatures is given below:
|
T/°C |
0 | 20 | 40 | 60 | 80 |
|
105 X K /S-1 |
0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and 1/T and calculate the values of A and Ea.
Predict the rate constant at 30 º and 50 ºC.
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 2: For the reaction: 2A + B → A2B the rate = k[A][B]2with k= 2.0 x 10-6mol-2L2s-1. Calculate....
Comments
oohhhooo AHhhaa
How come we halved the concentration of A to find the new one of B
Ohho
Second part of this answer\'s method is correct but answer is wrong...it will be = 3.888*10^-9 mol-2L2s-1