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nekit [7.7K]
3 years ago
8

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Chemistry
1 answer:
Masteriza [31]3 years ago
8 0

Answer:

\large \boxed{\text{-827.4 kJ}}

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + <u>2H₂O(g</u>)  ⟶ 2PbO(s) + <u>2H₂S(g</u>);  ∆H =  208.6 kJ

6. <u>2H₂S(g)</u> + O₂(g)        ⟶ <u>2S(s</u>)     + <u>2H₂O(g)</u> ; ∆H = -442.4 kJ

<u>7</u><u>. </u><u>2S(s)</u><u>      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ </u>

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}

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A typical excersise of solution.

It is more confortable to make a table for this.

                |   masss  |  volume  |  mol

solute       |                |                |          

solvent     |                |                |  

solution    |                |                |

Let's complete, what we have.

                 |   masss  |  volume  |  mol

solute       |  10.8g     |                |          

solvent     |                |  133 mL   |  

solution    |                |  137 mL    |

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From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

                 |   masss  |  volume  |  mol

solute       |   10.8g     |                |   0.110 mol      

solvent     |   133g      |  133 mL   |  7.39 mol

solution    |   143.8g   |  137 mL   | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles  or  solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

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