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Cloud [144]
3 years ago
15

Given that the initial rate constant is 0.0191 s-1 at an initial temperature of 24°C, what would the rate constant be at a tempe

rature of 120°C for a reaction with an activation energy of 43.4 kJ/mol?
Physics
1 answer:
wolverine [178]3 years ago
6 0

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

             ln \frac{K_{2}}{K_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

Since, it is given that K_{1} is 0.0191 s^{-1}, T_{1} is 24 degree celsius equals 297 K, T_{2} is 120 degree celsius equals 393 K, and E_{a} is 43.4 kJ/mol or 43400 J/mol.

Therefore, putting the value in the above formula as follows.

               ln \frac{K_{2}}{K_{1}} = \frac{-43400 J/mol}{8.314 j/mol K}[\frac{1}{393} - \frac{1}{297}]

                ln \frac{K_{2}}{0.0191 s^{-1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

                      K_{2} = 1.40 s^{-1}

Thus, we can conclude that rate constant be at a temperature of 120°C for a reaction with an activation energy of 43.4 kJ/mol is 1.40 s^{-1}.

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serg [7]

Answer:

Incomplete question:

Refer to the temperature versus time graph when answering the questions in Parts C through F. A system consists of 250 of water. The system, originally at = 21.0 , is placed in a freezer, where energy is removed from it in the form of heat at a constant rate. The figure shows how the temperature of the system takes to drop to, after which the water freezes. Once the freezing is complete, the temperature of the resulting ice continues to drop, reaching temperature after an hour. The following specific heat and latent heat values for water may be helpful.

specific heat of ice (at ) = 2.10 J/g K

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specific heat of water (at ) = 4.186 J/g K

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Answers:

Qtotal = 237775 J

Explanation:

To solve this exercise it is necessary to know that if the system is a single phase in which there is a temperature change or if it is a phase change at a single temperature. In the first case, the following formula would be used to calculate the amount of heat:

Q₁ = mCpΔT

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m is the mass = 250 g

Cp is the specific heat of ice = 2.1 J/g K

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In this case the amount of energy is

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Substituting:

Q₂ = 250*333.7 = 83425 J

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Qtotal = 154350+83425=237775 J

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Answer:

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