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2 years ago

Calculate the AH for the reaction C₂H4 (9) + H₂ (9)→ C₂H6 (g), from the following:

Chemistry

1 answer:

givi [52]2 years ago

5 0

The AH for the reaction is -137 kJ

<h3>What is Enthalpy of Formation ?</h3>

Enthalpy of formation is the energy required by reactants to convert into products . It value can be positive or negative depending on the type of reaction, Endothermic or Exothermic.

In the question 3 step reaction is given

C₂H4 (g) + 3 O₂ (g) → 2 CO₂ (g) + 2 H₂O (1) AH = - 1411. KJ

2 C₂H6 (g) + 7 O₂ (g) → 4 CO₂ (g) + 6 H₂O (1) AH = - 3120. KJ

2 H₂ (g) + O₂ (g) → 2 H₂O (1) AH = -571.6 kJ

To determine the enthalpy of formation for

C₂H4 (9) + H₂ (9)→ C₂H6 (g)

Adding reaction 1 and (1/2) 3 will give

C₂H4 (g) + 7/2 O₂ (g) + H₂ (g) ----> 2 CO₂ (g) + 3H₂O AH = -1696.8

For the second reaction

We will reverse the reaction and multiply it by 1/2

2 CO₂ (g) + 3 H₂O (1) --> C₂H6 (g) + 7/2 O₂ (g) AH = +1560

Adding the two new reactions.

C₂H4 (9) + H₂ (9)→ C₂H6 (g) AH = -137 kJ

Therefore The AH for the reaction is -137 kJ

To know more about Enthalpy of Formation

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Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

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g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

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1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

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