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The rate of a reaction is doubled for every 10°C rise in temperature. The increase in reaction rate as a result of temperature rise from 10°C to 100°C is?
Correct answer is '512'. Can you explain this answer?
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The rate of a reaction is doubled for every 10°C rise in temperatu...
The rate of the reaction is doubled for every 10oC rise in Temperature.
So the increase in reaction rate as a result of rise in temperature from 10oC to 100oC is 29 = 512 times
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The rate of a reaction is doubled for every 10°C rise in temperatu...
°C increase in temperature. This is known as the Arrhenius equation, which describes the relationship between the rate constant (k) and temperature (T):
k = A * e^(-Ea/RT)
where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
When the temperature increases by 10°C, the value of T in Kelvin increases by 10°C + 273.15 = 283.15 K. Thus, we can write:
k2/k1 = 2 = e^((Ea/R)*(1/T1 - 1/T2))
where k1 and T1 are the rate constant and temperature at the initial conditions, and k2 and T2 are the rate constant and temperature at the new conditions.
Solving for Ea, we get:
Ea = R * ln(2) * (T2*T1)/(T2 - T1)
Substituting the values, we get:
Ea = 8.314 J/mol*K * ln(2) * (293.15 K * 303.15 K)/(303.15 K - 293.15 K) = 54.6 kJ/mol
Therefore, the activation energy for this reaction is 54.6 kJ/mol.
k = A * e^(-Ea/RT)
where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
When the temperature increases by 10°C, the value of T in Kelvin increases by 10°C + 273.15 = 283.15 K. Thus, we can write:
k2/k1 = 2 = e^((Ea/R)*(1/T1 - 1/T2))
where k1 and T1 are the rate constant and temperature at the initial conditions, and k2 and T2 are the rate constant and temperature at the new conditions.
Solving for Ea, we get:
Ea = R * ln(2) * (T2*T1)/(T2 - T1)
Substituting the values, we get:
Ea = 8.314 J/mol*K * ln(2) * (293.15 K * 303.15 K)/(303.15 K - 293.15 K) = 54.6 kJ/mol
Therefore, the activation energy for this reaction is 54.6 kJ/mol.
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The rate of a reaction is doubled for every 10°C rise in temperature. The increase in reaction rate as a result of temperature rise from 10°C to 100°C is?Correct answer is '512'. Can you explain this answer?
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