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In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become
- a)64 times
- b)128 times
- c)256 times
- d)512 times
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
In a zero-order reaction for every 10 rise of temperature, the rate is...
For rise in temperature, n=1
Therefore rate = 2^n = 2^1 =2
When temperature is increased from 10 degree c to 100 degree c so, change in temperature = 100 -10 = 90 degree c , therefore n = 9 , therefore rate = 2^9 = 512 times...ans...
$$Hope it's help...$$
Therefore rate = 2^n = 2^1 =2
When temperature is increased from 10 degree c to 100 degree c so, change in temperature = 100 -10 = 90 degree c , therefore n = 9 , therefore rate = 2^9 = 512 times...ans...
$$Hope it's help...$$
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In a zero-order reaction for every 10 rise of temperature, the rate is...
Zero-order reaction and effect of temperature on rate
A zero-order reaction is a chemical reaction where the rate of reaction is independent of the concentration of the reactant(s). In other words, the rate of reaction remains constant throughout the reaction. However, the rate of reaction can be affected by temperature.
Effect of temperature on rate of reaction can be expressed by the Arrhenius equation:
k = Ae^(-Ea/RT)
Where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Doubling of rate with 10C rise in temperature
According to the problem statement, for every 10C rise in temperature, the rate of the reaction is doubled. Let's assume the initial rate of reaction at 10C is k1 and the rate of reaction at 20C is k2. Then, we can write:
k2/k1 = 2
Taking the natural logarithm of both sides, we get:
ln(k2/k1) = ln(2)
Using the Arrhenius equation, we can relate the rate constant at two different temperatures:
k2/k1 = (Ae^(-Ea/R(1/T2 - 1/T1))) / (Ae^(-Ea/R(1/T1 - 1/T1)))
Simplifying the equation, we get:
k2/k1 = e^(Ea/R * (1/T1 - 1/T2))
Substituting k2/k1 = 2 and T1 = 283 K (10C) and T2 = 373 K (100C), we get:
2 = e^(Ea/R * (1/283 - 1/373))
Solving for Ea, we get:
Ea = 53606 J/mol
Rate of reaction at 100C
Now, we can calculate the rate of reaction at 100C using the Arrhenius equation:
k = Ae^(-Ea/RT)
Substituting T = 373 K and the value of Ea we calculated earlier, we get:
k = Ae^(-53606/(8.314*373))
Simplifying the equation, we get:
k = Ae^(-17.58)
Since the rate of reaction is doubled for every 10C rise in temperature, the rate of reaction at 100C will be:
k' = 2^9 * k
k' = 512k
Therefore, the correct answer is option D, i.e., the rate of the reaction will become 512 times at 100C.
A zero-order reaction is a chemical reaction where the rate of reaction is independent of the concentration of the reactant(s). In other words, the rate of reaction remains constant throughout the reaction. However, the rate of reaction can be affected by temperature.
Effect of temperature on rate of reaction can be expressed by the Arrhenius equation:
k = Ae^(-Ea/RT)
Where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Doubling of rate with 10C rise in temperature
According to the problem statement, for every 10C rise in temperature, the rate of the reaction is doubled. Let's assume the initial rate of reaction at 10C is k1 and the rate of reaction at 20C is k2. Then, we can write:
k2/k1 = 2
Taking the natural logarithm of both sides, we get:
ln(k2/k1) = ln(2)
Using the Arrhenius equation, we can relate the rate constant at two different temperatures:
k2/k1 = (Ae^(-Ea/R(1/T2 - 1/T1))) / (Ae^(-Ea/R(1/T1 - 1/T1)))
Simplifying the equation, we get:
k2/k1 = e^(Ea/R * (1/T1 - 1/T2))
Substituting k2/k1 = 2 and T1 = 283 K (10C) and T2 = 373 K (100C), we get:
2 = e^(Ea/R * (1/283 - 1/373))
Solving for Ea, we get:
Ea = 53606 J/mol
Rate of reaction at 100C
Now, we can calculate the rate of reaction at 100C using the Arrhenius equation:
k = Ae^(-Ea/RT)
Substituting T = 373 K and the value of Ea we calculated earlier, we get:
k = Ae^(-53606/(8.314*373))
Simplifying the equation, we get:
k = Ae^(-17.58)
Since the rate of reaction is doubled for every 10C rise in temperature, the rate of reaction at 100C will be:
k' = 2^9 * k
k' = 512k
Therefore, the correct answer is option D, i.e., the rate of the reaction will become 512 times at 100C.
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In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer?.
In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a zero-order reaction for every 10 rise of temperature, the rate is doubled. If the temperature is increased from 10C to 100C, the rate of the reaction will becomea)64 timesb)128 timesc)256 timesd)512 timesCorrect answer is option 'D'. Can you explain this answer?.
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