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2 years ago

When aqueous magnesium chloride reacts with liquid bromine

Chemistry

1 answer:

Bingel [31]2 years ago

3 0

Word equation: magnesium chloride + bromine -> magnesium bromide + chlorine
balance equation: MgCl2 + 2Br -> MgBr2 + 2Cl

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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!

4 0

3 years ago

Who can answer these?

sesenic [268]

For the first one.. the Reactant side is Zn+2HCl produces Zncl2 and H2... and it is balanced too

6 0

3 years ago

PLEASE HELP! Class is almost ending, I don't want to repeat.

nikklg [1K]

ANSWERS:

1. $2KClO_{3}$ ⇒ 2KCl + $3O_{2}$

2. $H_{2} C_{2} O_{4} + 2NaOH$ ⇒ $Na_{2} C_{2} O_{4} + 2 H_{2} O$

3. $2 C_{10} H_{22} + 31O_{2}$ ⇒ $20C O_{2} + 22 H_{2} O$

4. $4 H_{2} + Fe_{3} O_{4}$ ⇒ $3Fe +4 H_{2} O$

5. $4P + 5 O_{2}$ ⇒ $2 P_{2} O_{5}$

I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)