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Rus_ich [418]
3 years ago
13

[Standard Enthalpy of Formation]

Chemistry
2 answers:
Pie3 years ago
7 0
This is the answer using hess’ law. you need to make the 3 equations equal the final one

SpyIntel [72]3 years ago
3 0

Answer:

5. ΔH = 238.7 kJ; 6 ΔH, = -1504.6 kJ

Step-by-step explanation:

Question 5:

We have three equations:  

(I)  CH₃OH + ³/₂O₂ ⟶ CO₂ + 2H₂O;  ΔH = -726.4 kJ

(II)  C + O₂ ⟶ CO₂;                              ΔH = -393.5 kJ

(III) H₂ + ½O₂ ⟶ H₂O;                         ΔH = -285.8 kJ

From these, we must devise the target equation:  

(IV) C + 2H₂ + ½ O₂ → CH₃OH; ΔH = ?  

The target equation has 1C on the left, so you rewrite Equation (II)..  

(V) C + O₂ ⟶ CO₂;                               ΔH = -393.5 kJ

Equation (V) has 1CO₂ on the right, and that is not in the target equation.  

You need an equation with ½O₂ on the left, so you <em>reverse Equation (I).  </em>

When you reverse an equation, you r<em>everse the sign</em> of its ΔH.  

(VI) CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂;  ΔH = 726.4 kJ

Equation (VI) has 2H₂O on the left, and that is not in the target equation.

You need an equation with 2H₂O on the right. Double Equation (III).

When you <u>double an equation, you double its ΔH</u>.  

(VII) 2H₂ + O₂ ⟶ 2H₂O;                        ΔH = -571.6 kJ

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear <em>on opposite sides</em> of the reaction arrows.

When you add equations, you add their ΔH values.

We get the target equation (IV)  

(V)   C + O₂ ⟶ CO₂;                              ΔH = -393.5 kJ

(VI)  CO₂ + 2H₂O ⟶ CH₃OH + ³/₂O₂ ;  ΔH =  726.4 kJ

<u>(VII</u><u>) 2H₂ + O₂ ⟶ 2H₂O;          </u>              <u>ΔH = -571.6 kJ </u>

(IV)  C + 2H₂ + ½ O₂ → CH₃OH;             ΔH = -238.7 kJ  

Question 6

We have three equations:  

(I) Mg + ½O₂ → MgO;  ΔH = -601.7 kJ

(II) Mg + S ⟶ MgS;      ΔH = -598.0 kJ

(III) S + O₂ ⟶ SO₂;       ΔH = -296.8 kJ

From these, we must devise the target equation:  

(IV) 3Mg + SO₂ → MgS + 2MgO; ΔH = ?  

The target equation has 1SO₂ on the left, so you reverse Equation(III).

(V) SO₂ ⟶ S + O₂;         ΔH = 296.8 kJ

Equation (V) has 1S on the right, and that is not in the target equation.  

You need an equation with 1S on the left, so you rewrite Equation (II).  

(VI) Mg + S ⟶ MgS;        ΔH = -598.0 kJ

Equation (V) has 1O₂ on the right, and that is not in the target equation.

You need an equation with 1O₂ on the left. Double Equation (I).

(VII) 2Mg + O₂⟶ 2MgO ;  ΔH = -1203.4 kJ  

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

We get the target equation (IV)  

(V)   SO₂ ⟶ S + O₂;                    ΔH =    296.8 kJ

(VI)   Mg + S ⟶ MgS;                  ΔH = -  598.0 kJ

<u>(VII)</u><u> 2Mg + O₂ ⟶ 2MgO;          </u>  <u>ΔH = -1203.4 kJ </u>

(IV)  3Mg + SO₂ → MgS + 2MgO; ΔH = -1504.6 kJ

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Answer:

             %age Yield  =  51.45 %

Solution:

Step 1: Convert Kg into g

68.5 Kg CO  =  68500 g CO

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Step 2: Find out Limiting reactant;

The Balance Chemical Equation is as follow;

                                 CO  +  2 H₂    →    CH₃OH

According to Equation,

                   28 g (1 mol) CO reacts with  =  4 g (2 mol) of H₂

So,

                    68500 g CO will react with  =  X g of H₂

Solving for X,

                    X  =  (68500 g × 4 g) ÷ 28 g

                    X  =  9785 g of H₂

It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.

Step 3: Calculate Theoretical Yield

According to equation,

            4 g (2 mol) H₂ reacts to produce  =  32 g (1 mol) Methanol

So,

                          8600 g H₂ will produce  =  X g of CH₃OH

Solving for X,

                    X  =  (8600 g × 32 g) ÷ 4 g

                     X =  68800 g of CH₃OH

Step 4: Calculate %age Yield

                     %age Yield  =  Actual Yield ÷ Theoretical Yield × 100

Putting Values,

                     %age Yield  =  3.54 × 10⁴ g ÷ 68800 g × 100

                     %age Yield  =  51.45 %


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