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Enzymes work at optimum temperature. Ove...

Enzymes work at optimum temperature. Over a range `0-40^(@)C`, what would happen to the rate of enzyme controlled reactions for every `10^(@)C` rise in temperature ?

A

The rate doubles itself

B

Decreases by half

C

No effect

D

First increases than decreases

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The correct Answer is:
To answer the question regarding the effect of temperature on the rate of enzyme-controlled reactions, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Enzymes**: Enzymes are biological catalysts that speed up chemical reactions in living organisms. They are sensitive to temperature changes. 2. **Optimum Temperature**: Each enzyme has an optimum temperature at which it functions most efficiently. This temperature allows the enzyme to have peak activity. 3. **Temperature Range**: The question specifies a temperature range of 0 to 40 degrees Celsius. As the temperature increases within this range, the activity of the enzyme also increases. 4. **Rate of Reaction**: Generally, for many enzymes, the rate of reaction approximately doubles with every 10 degrees Celsius increase in temperature, up to a certain point (the optimum temperature). 5. **Graphical Representation**: If we were to plot a graph of enzyme activity against temperature, we would see an initial increase in the rate of reaction with rising temperature, reaching a peak at the optimum temperature, and then a decline as the temperature continues to rise beyond this point. 6. **Conclusion**: Therefore, for every 10 degrees Celsius rise in temperature from 0 to 40 degrees Celsius, the rate of enzyme-controlled reactions would approximately double until the optimum temperature is reached. ### Final Answer: The rate of enzyme-controlled reactions would double for every 10 degrees Celsius rise in temperature, up to the optimum temperature. ---
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The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) For a reaction E_(a) = 0 and k = 3.2 xx 10^(8)s^(-1) at 325 K . The value of k at 335 K would be

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) For which of the following reactions k_(310)//k_(300) would be maximum?

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