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spayn [35]
3 years ago
14

The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)

are +112 and +121.0 kJ per mol of atoms respectively, calculate the average enthalpy of the Br-Cl bond in BrCl(g).
Chemistry
1 answer:
natita [175]3 years ago
8 0

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

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Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

5 0
3 years ago
In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water,
Dahasolnce [82]

Answer:

<h2>15 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

But from the question

volume = final volume of water - initial volume of water

volume = 165 - 150 = 15 mL

We have

density =  \frac{225}{15}  = 15 \\

We have the final answer as

<h3>15 g/mL</h3>

Hope this helps you

5 0
2 years ago
Fritz-Haber process
maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0
3 years ago
The two main products of the combustion of gasoline in an automobile engine are
kozerog [31]
There is an excess of oxygen, the two main products should be carbon dioxide gas and water vapor. If the oxygen supply is limited, the fuel will undergo incomplete combustion and produce carbon monoxide, water, and sometimes carbon.

Hope this helped! :)
4 0
3 years ago
Read 2 more answers
Pls help me<br> what bond does not burn
patriot [66]

Answer:

Iconic bonds don't burn easily

Explanation:

Covalent bonds are non metals. Covalent bond (sharing), low temp, low temp, burn easily, poor, polar covalent is good and non-polar covalent is bad.

Ionic - metals and nonmetals, ionic bond is when electrons are gained or lost, high temp, high temp, doesn't burn easily, good, good

5 0
3 years ago
Read 2 more answers
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