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3 years ago

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Carbon monoxide and chlorine combine in an equilibrium reaction to produce the highly toxic product, phosgene (COCl2 ) CO(g) Cl2

(g) COCl2 (g) If the equilibrium constant for this reaction is Kc = 248, predict, if possible, what will happen when the reactants and product are combined with the concentrations shown. [CO] = [Cl2] = 0.010 M; [COCl2] = 0.070 M

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

Reaction will proceed towards forward direction.

Explanation:

Balanced reaction: $CO+Cl_{2}\rightleftharpoons COCl_{2}$

Reaction quotient ( $Q_{c}$ ) for this reaction is represented as:

$Q_{c}=\frac{[COCl_{2}]}{[CO][Cl_{2}]}$

Here, [CO] = [ $Cl_{2}$ ] = 0.010 M and [ $COCl_{2}$ ] = 0.070 M

So, $Q_{c}=\frac{0.070}{(0.010)^{2}}$ = 700

As $Q_{c}$ > $K_{c}$ therefore reaction will proceed towards forward direction i.e. more $COCl_{2}$ will be produced.

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A 17.6-g sample of ammonium carbonate contains ________ mol of ammonium ions.

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These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate.

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

17.6 gNH4CO3

Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol

17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4

Now just take the molar mass we found to convert that amount into moles!

4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4

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Which will provide the most direct measure of pH by measuring the hydrogen ion concentration of a solution?

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1 year ago

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goblinko [34]

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3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago