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The rate of chemical reaction X is inversely proportional to the squar
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31 Aug 2022, 12:15
31 Aug 2022, 12:15
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Question Stats:
70% (01:34) correct
30% (02:23) wrong based on 10 sessions
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The rate of chemical reaction X is inversely proportional to the square of the rate of chemical reaction Y. If the rate of chemical reaction Y increases by 100%, by what per cent does the rate of chemical reaction X decrease?
A. 50%
B. 60%
C. 70%
D. 75%
E. 80%
A. 50%
B. 60%
C. 70%
D. 75%
E. 80%
Re: The rate of chemical reaction X is inversely proportional to the squar
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31 Aug 2022, 22:11
31 Aug 2022, 22:11
2
Expert Reply
Considering thet Y is squared and the increase is 100%. Then the decrease should be halved so 50% But because is squared we do have half of the half so in the end 25%
Overall decrease is 75%
Overall decrease is 75%
Re: The rate of chemical reaction X is inversely proportional to the squar
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31 Aug 2022, 23:30
31 Aug 2022, 23:30
1
Alternative Solution -
The rate of chemical reaction X is inversely proportional to the square of the rate of chemical reaction Y –
So, this can be represented as:
Rate of Chemical Reaction X=[1 / (Y^2)]
The rate of Chemical Reaction Y increases by 100% which implies, that Y + 100% of Y = 2Y
Since the rate of chemical reaction X and the rate of chemical reaction Y are inversely proportional, we can represent the new rate as:
Rate of Chemical Reaction X = [1 / (2Y)^2]
So, the decrease in the rate of chemical reaction X = [1 / (Y^2)] - [1 / (2Y)^2] = [3 / (4Y)^2]
Percentage decrease in the rate of chemical reaction = {[3 / (4Y)^2] / [1 / (Y^2)]} * 100 = 3/4 * 100 = 75 %
The rate of chemical reaction X is inversely proportional to the square of the rate of chemical reaction Y –
So, this can be represented as:
Rate of Chemical Reaction X=[1 / (Y^2)]
The rate of Chemical Reaction Y increases by 100% which implies, that Y + 100% of Y = 2Y
Since the rate of chemical reaction X and the rate of chemical reaction Y are inversely proportional, we can represent the new rate as:
Rate of Chemical Reaction X = [1 / (2Y)^2]
So, the decrease in the rate of chemical reaction X = [1 / (Y^2)] - [1 / (2Y)^2] = [3 / (4Y)^2]
Percentage decrease in the rate of chemical reaction = {[3 / (4Y)^2] / [1 / (Y^2)]} * 100 = 3/4 * 100 = 75 %
