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Sunny_sXe [5.5K]
3 years ago
10

Consider these reactions, where M represents a generic metal.

Chemistry
1 answer:
weeeeeb [17]3 years ago
5 0

Consider these reactions, where M represents a generic metal.  Determine ΔH₅.

  1. 2M (s) + 6HCl (aq) ⟶ 2MCl₃ (aq) + 3H₂ (g); ΔH₁ = −878.0 kJ
  2. HCl (g) ⟶ HCl (aq); ΔH₂ = −74.8 kJ
  3. H₂ (g) + Cl₂ (g) ⟶ 2HCl (g); ΔH₃ = −1845.0 kJ
  4. MCl₃ (s) ⟶ MCl₃ (aq); ΔH₄ = −497.0 kJ
  5. 2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s); ΔH₅ = ? kJ

We need to manipulate the equations on top to output the equation on the bottom.  Hess's Law states that the total enthalpy change of the desired reaction is the sum of all ethalpy changes:

ΔH_{overall}=H_{1}  +H_{2} +H_{3} +...+H_{n}.

We will need to arrange the equations so that the compounds/ elements that are not in the desired reaction will cancel out.  This can be done by reversing the reactions, and by multiplying or dividing a given equation by a factor.

For reactants in our desired reaction we have 2M (s) and 3Cl₂ (g) and their product is 2MCl₃ (s), so the compounds/ elements in the equations that will need to cancel are HCl (aq), H₂ (g), HCl (g), and MCl₃ (aq).

  • HCl (aq).  The HCl (aq)'s are on opposite sides of the arrow, but they have different coefficients.  To be able to cancel, you will need to multiply the second HCl (aq) by 6.  Just like in algebra, you have to multiply the entire equation, not just one variable.  Also, since you're multiplying the equation by 6, its enthalpy will also increase by a factor of 6.

2. 6(HCl (g) ⟶ HCl (aq)); ΔH₂ = 6(−74.8) kJ

2. 6HCl(g) ⟶ 6HCl (aq); ΔH₂ = -448.8 kJ

  • H₂ (g).  The equations with hydrogen gas are the first and the third.  The first hydrogen has a coefficient of 3, and the second, a coefficient of 1.  Multiply the second equation and its enthalpy by 3:

3. 3(H₂ (g) + Cl₂ (g) ⟶ 2HCl (g)); ΔH₃ = 3(−1845.0) kJ

3. 3H₂ (g) + 3Cl₂ (g) ⟶ 6HCl (g); ΔH₃ = -5535.0 kJ

  • HCl (g).  The equations containing HCl in a gaseous form are equations 2 and 3, which we already manipulated for other variables, but in the process we made kept the HCl (g)'s the same.

2. 6HCl(g) ⟶ 6HCl (aq); ΔH₂ = -448.8 kJ

3. 3H₂ (g) + 3Cl₂ (g) ⟶ 6HCl (g); ΔH₃ = -5535.0 kJ

  • MCl₃ (aq).  In equation 1, MCl₃ (aq) has a coefficient of 2.  Multiply equation 4 and its enthalpy by 2.  To cancel terms, you will need to reverse the equation.  This changes the sign on the enthalpy from negative to positive.

4. 2(MCl₃ (s) ⟶ MCl₃ (aq)); ΔH₄ = 2(−497.0) kJ

4. 2MCl₃ (s) ⟶ 2MCl₃ (aq); ΔH₄ = -994.0 kJ

4. 2MCl₃ (aq) ⟶ 2MCl₃ (s); ΔH₄ = 994.0 kJ

Now when you add all the products and the reactants, the resulting equation should be 2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s) (underlined terms cancel since they are on both sides):

2M (s) + 6HCl (aq) ⟶ <em>2MCl₃ (aq) + 3H₂ (g)</em>

6HCl(g) ⟶ <em>6HCl (aq)</em>

3H₂ (g) + 3Cl₂ (g) ⟶ <em>6HCl (g)</em>

2MCl₃ (aq) ⟶ <em>2MCl₃ (s)</em>

2M (s) + <u>6HCl (aq)</u> + <u>6HCl(g)</u> + <u>3H₂ (g)</u> + 3Cl₂ (g) + <u>2MCl₃ (aq)</u> ⟶ <em><u>2MCl₃ (aq) </u></em><em>+ </em><em><u>3H₂ (g) </u></em><em>+ </em><em><u>6HCl (aq)</u></em><em> + </em><em><u>6HCl (g) </u></em><em>+ 2MCl₃ (s)</em>

2M (s) + 3Cl₂ (g) ⟶ 2MCl₃ (s)

Finally, add the enthalpies:

−878.0 kJ + (-448.8 kJ) + (-5535.0 kJ) + 994.0 kJ = -5869 kJ

<h3>Answer:</h3>

ΔH₅ = -5869 kJ

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The moles of helium gas is calculated using the Ideal gas equation:

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Using P1/T1 = P2/T2

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The pressure at peak temperature is 259.3 kPa

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An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
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<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

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The balanced chemical equation is:

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Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

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The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)

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