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Elan Coil [88]
4 years ago
5

Using this reversible reaction, answer the following:

Chemistry
2 answers:
spin [16.1K]4 years ago
4 0
<span>-Products formed by the forward 
reaction may react with each other to regenerate the reactants. 
-When reactants are mixed, they will begin to react at a forward reaction rate particular to that chemical reaction. 
- As reactants are depleted and products are formed, however, the rate of the forward reaction begins to slow, and the rate of the reverse reaction begins to increase.</span>
Mamont248 [21]4 years ago
4 0
<h2>Question 1.</h2><h3>Answer:</h3>

The concentration of [N2O4] will decrease.

<h3>Explanation:</h3>
  • As the reverse reaction which is the formation of N2O4 is an exothermic reaction and the formation of NO2 is an endothermic reaction.
  • Hence heating will favor the forward reaction and stops the reverse reaction.
  • N2O4 --> 2NO2
  • So the concentration of N2O4 decreases.
<h2>Question: 2.</h2><h3>Answer:</h3>

The formation of the products is favored by the addition of heat.

<h3>Explanation:</h3>
  • As the reverse reaction which is the formation of N2O4 is an exothermic reaction and the formation of NO2 is an endothermic reaction.
  • So the addition of heat will favor endothermic reaction.
  • Hence the formation of NO2 will be favored which is a product.
<h2>Question no.4:</h2><h3>Answer:</h3>

The reverse reaction is an exothermic reaction.

N2O4 <-- 2NO2.

<h3>Explanation:</h3>
  • N2O4 which is a reactant in this reaction is colorless compound.
  • Its breakdown absorbs energy and its formation release energy in form of heat.
  • So the reverse reaction which releases energy is an exothermic reaction.
<h2>Question no. 4</h2><h3>Answer:</h3>

The correct option is D which is N2O4 <--> 2NO2, HR = -14 kcal.

<h3>Explanation:</h3>
  • Enthalpy is the change in energy in reactants and products.
  • It is actually the difference in energies of products and reactant.
  • As the enthalpy change from left to right, it means reactants have internal energy 14 kcal high than products and it will be released.
  • And in option D which is shown as HR = -14 kcal.

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Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution i
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y_{C_6H_{14}}=0.92

Explanation:

Hello,

At first, we compute liquid-phase molar fractions:

n_{C_6H_{14}}=463.96 g*\frac{1mol}{86g} =5.3949molC_6H_{14}\\n_{C_7H_{16}}=667.71 g*\frac{1mol}{100g} =6.6771molC_7H_{16}\\x_{C_6H_{14}}=\frac{5.3949}{5.3949+6.6771} =0.447\\x_{C_7H_{16}}=1-x_{C_6H_{14}}=0.553

Now, by means of the fugacity concept, for hexane, for instance, we have:

f_{C_6H_{14}}^V=f_{C_6H_{14}}^L\\y_{C_6H_{14}}p_T=x_{C_6H_{14}}p_{C_6H_{14}}

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

p_T=x_{C_6H_{14}}p_{C_6H_{14}}+x_{C_7H_{16}}p_{C_7H_{16}}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

y_{C_6H_{14}}=\frac{x_{C_6H_{14}}p_{C_6H_{14}}}{p_T}=\frac{0.447*0.198946 atm}{0.096622atm} \\y_{C_6H_{14}}=0.92

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Percent Composition =

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S = 56.8%

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