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finlep [7]
2 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]2 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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If a gas sample has a pressure of 30.7 kPa at 0.00*C, by how much does the temperature have to decrease to lower the pressure to
Scrat [10]

Answer:

                      252.68 K  or   -20.46 °C

Explanation:

                    According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".

Mathematically,

                                              P₁ / T₁  =  P₂ / T₂   ---- (1)

Data Given:

                  P₁  =  30.7 kPa

                  T₁  =  0.00 °C  =  273.15 K

                  P₂  =  28.4 kPa

                  T₂  =  <u>???</u>

Solving equation for T₂,

                  T₂  =  P₂ T₁ / P₁

Putting values,

                  T₂  =  28.4 kPa × 273.15 K / 30.7 kPa

                  T₂  =  252.68 K  or   -20.46 °C

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