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leva [86]
3 years ago
7

For a reversible reaction, what would a large equilibrium constant indicate?

Chemistry
1 answer:
siniylev [52]3 years ago
8 0

(D) At equilibrium, the concentration of the products will be much higher than the concentration of the reactants.

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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

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Answer:

Electron transport is the process by NADH + H+ and FADH2 are converted to NAD+ and FAD, donating electrons and hydrogen ions to oxygen

Explanation:

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HELP . which kind of tides causes the highest and lowest tides ?
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