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Murrr4er [49]
3 years ago
9

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated wit

h oxides as a catalyst. For the following reaction, equilibrium constants (Kp values) are seen below as a function of temperature.
N2(g) + 3 H2(g) reverse reaction arrow 2 NH3(g)

300°C 4.34 multiplied by 10-3 atm-2
500°C 1.45 multiplied by 10-5 atm-2
600°C 2.25 multiplied by 10-6 atm-2
This is an exothermic reaction, can you explain why it is exothermic and not endothermic?
Chemistry
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

Explanation:

The reaction of production of ammonia is as follows

N₂ + 3H₂ ⇄ 2NH₃

From the data given , it is clear that

equilibrium constant decreases as temperature increases . In other words

at higher temperature less and less of ammonia is produced .

According to Le Chatelier's Principle , reaction tends to go in the direction so that the effect of external change is minimized. Since the increase in temperature reduces the forward reaction , that means reaction is exothermic .

So , reduction in reaction , helps reduce the production of heat to reduce the effect of temperature increase.

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C=h/[M(T2-T1)] solve for T1
vodka [1.7K]

Explanation:

This is expressing as a subject of a formula of a specie.

This equation is used in solving for specific heat capacity of a material;

     equation:

         c = \frac{h}{M(T2 - T1)}

First cross multiply:

  c[M(T2 - T1)]  = h

Then multiply both sides by \frac{1}{c } :

\frac{1}{c }  x    c[M(T2 - T1)]   = \frac{1}{c }  x h

        M(T2 - T1) = \frac{h}{c}

Multiply both sides by \frac{1}{M}

\frac{1}{M}   x   M(T2 - T1)  = \frac{1}{M} x  \frac{h}{c}

    T2 - T1 = \frac{h}{Mc}

Now rearrange to produce;

    T1 = T2 + \frac{h}{Mc}

learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

7 0
4 years ago
An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.507 V

Now put all the given values in the above equation, we get:

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
Which formula equation shows a reversible reaction?
pishuonlain [190]

Answer:2NaF is the correct one. It’s a simple combination and can be be split with relative ease

Explanation:

3 0
3 years ago
Read 2 more answers
The cells that allow your bones to move contain bundles of long cylinders. These cells have alternating light and dark bands cal
poizon [28]

Answer:

<h2>Actin and myosin.</h2>

Explanation:

The cells that allow your bones to move, the movement of thick (myosin) and thin (actin) filaments during contraction .

During a contraction thick and thin filaments do not shorten but increase their overlap  of each other.

Thin filaments slide past thick filaments extending more deeply into the A band.

The I bands and H bands decrease in lenght as Z discs are come closer together .

Sarcomere represents area between two Z disc, so the sarcomere gets smaller during a contraction .

8 0
3 years ago
The reaction of pyrrole with bromine forms predominantly __________. View Available Hint(s) The reaction of pyrrole with bromine
xenn [34]

Answer:

a) 2-bromopyrrole

Explanation:

Our options for this questions are:

a) 2-bromopyrrole

b) 2,3-dibromopyrrole

c) N-bromopyrrole

d) 3-bromopyrrole

To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.

The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the Br_2 and FeBr_3. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).

I hope it helps!

5 0
3 years ago
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