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chetanojha
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Can somebody help me to understand solution for this question or direct me to link in this forum if discussed earlier?
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italian7745
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This question is basically an application question of indirect proportions...
Let R1 be the rate and A1 and B1 be the concentrations..
R1=K1.A1^2=K2/B1...EQUATION 1)...
Now concentration of B1 is increased by 100% so B2=2B1...
R2=K3.A2^2=K4/B2...EQUATION 2)...
We know that rate is inversely proportional to concentration of B..we have increased the quantity of B ...So the rate shall decrease...which leads to the application of indirect proportion...
To maintain the same rate , R1=R2...
R1/R2 = K1.A1^2/K3.A2^2 = K4/B2/K2/B1...EQUATION 3).
After solving this equation...
A1^2.B2=A2^2.B1
A2^2/A1^2=2
A2/A1=(Square root) 2
A2= 1.414 A1..
So concentration is approx. increased by 40%...
Let R1 be the rate and A1 and B1 be the concentrations..
R1=K1.A1^2=K2/B1...EQUATION 1)...
Now concentration of B1 is increased by 100% so B2=2B1...
R2=K3.A2^2=K4/B2...EQUATION 2)...
We know that rate is inversely proportional to concentration of B..we have increased the quantity of B ...So the rate shall decrease...which leads to the application of indirect proportion...
To maintain the same rate , R1=R2...
R1/R2 = K1.A1^2/K3.A2^2 = K4/B2/K2/B1...EQUATION 3).
After solving this equation...
A1^2.B2=A2^2.B1
A2^2/A1^2=2
A2/A1=(Square root) 2
A2= 1.414 A1..
So concentration is approx. increased by 40%...
