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tamaranim1 [39]
2 years ago
5

For an experiment, you place 15 g of ice with a temperature of -10 C into a cup and label

Chemistry
2 answers:
melamori03 [73]2 years ago
6 0
The correct answer is c. Temperature is the average kinetic energy of a sample so if two samples have the same temperature they will also have the same average kinetic energy. I hope this helps. Let me know if anything is unclear.
Darina [25.2K]2 years ago
5 0

Average kinetic energy is given by the formula:

KE(average) = 3/2*k*T

where k = Boltzmann constant = 1.38*10⁻²³ J/K

T = temperature in Kelvin (K)

Thus, the average KE is dependent only on temperature and independent of mass.

Since the temp in Cup A and B remains the same, the average KE will also be the same.

Ans C)


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A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
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A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

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From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

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             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

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R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

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