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finlep [7]
3 years ago
10

For the reaction where Δn=−1Δn=−1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactant

s. For the reaction where Δn=0Δn=0 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants. For the reaction where Δn=+1Δn=+1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants.
Chemistry
1 answer:
Alex787 [66]3 years ago
8 0

Answer:

Explanation:

In general, an increase in pressure (decrease in volume) favors the net reaction  that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases  the total number of moles of gases.

Δn= b - a

Δn=  moles of gaseous products - moles of gaseous reactants

Therefore, <u>after the increase in volume</u>:

  • If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
  • If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
  • Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.

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<u>Explanation:</u>

We are given:

Molality of solution = 1 m

This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of solute = 1 moles

Total moles = [1 + 55.56] = 56.56 moles

Putting values in above equation, we get:

\chi_{(solute)}=\frac{1}{56.56}=0.0177

The equation used to calculate relative lowering of vapor pressure follows:

\frac{p^o-p_s}{p^o}=i\times \chi_{solute}

where,

\frac{p^o-p_s}{p^o} = relative lowering in vapor pressure = 0.734 mmHg

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p^o = vapor pressure of pure water = 23.76 torr

Putting values in above equation, we get:

\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74

Hence, the Van't Hoff factor for KCl is 1.74

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