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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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Explanation:

As the units for concentration are not specified, I'll respond using g/mL.

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4 0
3 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


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In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:  

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)

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