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3 years ago

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A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the h

igher temperature to f at the lower temperature?

1 answer:

Kazeer [188]3 years ago

8 0

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

$f=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$f_1$ = rate constant at 525K

$K_2$ = rate constant at 545K

$Ea$ = activation energy for the reaction = 185kJ/mol= 185000J/mol (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 525 K

$T_2$ = final temperature = 545 K

Now put all the given values in this formula, we get

$\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]$

$\log (\frac{f_2}{f_1})=0.6754$

$(\frac{f_2}{f_1})=4.736$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736

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Given:

Inductance, L = 150 mH

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$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

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e). Phase angle between current and the generator voltage is given by:

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$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

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Explanation :

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