Which of the following characterizes a reaction at equilibrium?

Determine the equilibrium constant, Keq, at 25°C for the reaction

adelina 88 [10]

Explanation:

The given chemical reaction is:

$2Br^- (aq) + I_2(s) Br_2(l) + 2I^- (aq)$

$E^ocell=oxidation potential of anode + reduction potential of cathode\\$

The relation between Eo cell and Keq is shown below:

$deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell$

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

$E^ocell= (-1.07+0.53)V\\=-0.54V$

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

$ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3$

Answer:

Keq=6.13x10^33

3 0

3 years ago

You need to make an aqueous solution of 0.182 M aluminum sulfate for an experiment in lab, using a 250 mL volumetric flask. How

Ksju [112]

Answer:

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask

The molar mass of Aluminum Sulfate = 342.15 g/mol

Using the molarity formula:-

Molarity = Number of moles/Volume of solution in a liter

Number of moles = Given weight/ molar mass

Molarity = (Given weight/ molar mass)/Volume of solution in liter

0.187 M = (Given weight/342.15 g/mol)/0.250 L

Given weight = 15.99 g

8 0

3 years ago

An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz =106sâ€“1). Calculate the w

kap26 [50]

Answer:

$Wavelength=2.899\ m$

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength$

c is the speed of light having value $2.998\times 10^8\ m/s$

Given, Frequency = 103.4 MHz = $103.4\times 10^{6}\ Hz$ ( as 1 MHz = 10⁶ Hz)

Thus, Wavelength is:

$Wavelength=\frac{c}{Frequency}$

$Wavelength=\frac{2.998\times 10^8}{103.4\times 10^{6}}\ m$

$Wavelength=2.899\ m$

5 0

3 years ago

Help me out please!!

malfutka [58]

Answer: $6.48*10^{-2}$

Explanation:

alright, dawg, lets get this bread. CHEMISTRY? OH YEAH I LOVE CHEMISTRY.

what is a mol? do you know who avogadro is? sounds like avocado. free shavocado. ok so you MUST REMEMBER THIS NUMBER PLEASE.

please remember this number and commit it to your memory: avogadros number

$6.02 * 10^{23}$

this is how much a mole is. you know how a pair is 2 and a dozen is 12? ok so a mole is $6.02 * 10^{23}$ it is confusing at first but hopefully this helps you to understand.

now that we understand this..... lets perform this calculation with a calculator

$(3.90 * 10^{22}) / (6.02 * 10^{23})$

notice i divide the question by the avogadros number to find out how many moles are in the number. ok but listen... it gets into a tough area here... because HOW ARE WE TO DIVIDE SUCH A HUMONGOUS NUMBER BY ANOTHER HUMONGOUS NUMBER?!?!?

its easy, its cake, just listen this is how you do it. only focus on the numbers NOT the 10 exponential ones. so just 3.90 and 6.02 ok? lets divide these two numbers 3.90 / 6.02 and we get 0.6478... how interesting... ok now lets deal with the exponents of 10. notice that we are DIVIDING these numbers so think of it as subtracting the exponents of ten..... 22 minus 23 equals -1

so we have $0.6478 * 10^{-1}$

now this negative 1 thing is annoying so lets just make it to the power of 0

$0.06478 * 10^{0}$

and anything to the power of 0 just becomes 1.

0.06478

so this is our answer but keep in mind we need 3 sig figs. if we round then we get 0.0648

put this into scientific notation we get $6.48*10^{-2}$

5 0

3 years ago

calcium reacts with fluorine to produce calcium fluoride. how does oxidation and reduction take place in this reaction?

Assoli18 [71]

The oxidation is occurring on Calcium ions as it release one electron and reduction will be occurring on fluorine ion as it accepts one electron.

<u>Explanation:</u>

An element will undergo oxidation and form a positive ion on releasing one or more electrons from its valence shell. While reduction is occurred in a chemical reaction, then the element will be forming a negative ion with the acceptance of one or more electrons in its valence shell.

So in the given process of calcium fluoride, the one electron from the valence shell of calcium will be released making it as $c a^{+}$ ions and this is termed as oxidation process. This one electron will be getting accepted by the fluorine ion and thus it will convert to $F^{-}$ ions. This process of acceptance of electrons is termed as reduction.

3 0

3 years ago

Which of the following characterizes a reaction at equilibrium?​

Which of the following characterizes a reaction at equilibrium?