Please anewer 12 question (b) part !
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this is the answer of the question number 12
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priyankapawar:
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Balancing the equation: ion electron method
2 (Mn O4)⁻ (aq) + 5 SO₂ (g) ⇒ 2 Mn²⁺ (aq) + 5 (HSO4)⁻ (aq) + e⁻
=> Acidic medium: So Add H⁺ ions and H₂O available in the medium.
2 MnO₄⁻ + 5 SO₂ + 2 H₂O + H⁺ => 2 Mn²⁺ + 5 HSO₄⁻
The balancing process is longer with intermediate half reactions and intermediate products. We can use some coefficients a , b, c and then balance each element on both sides. It is more like algebra.
Alternatively, we write the half reactions for oxidation and reduction of the important elements involved. Then we add them together while balancing them. Further we remove the common compounds or ions or charges on both sides. We are left with the balanced final equation.
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a) Le Chatelier principle:
For a chemical reaction in a chemical equilibrium, if any change occurs in pressure, temperature or concentration then the system will favour the equilibrium towards the side that opposes the change.
The effect will be to shift the equilibrium towards the side that minimizes (absorbs) the changes. A new equilibrium is established. Kc or Kp will remain constant at the given temperature.
b) Application of Le Chatelier's principle
2 H₂ (g) + CO (g) ⇔ CH₃ OH (g)
Kc = [ CH₃ OH ] / { [ H₂ ]² * [ CO ] } = constant at a Temperature
i) H2 increased.
After the equilibrium is established, [ H₂ ] increases while others are constant, then equilibrium shifts to the right. So [ H₂ ] will decrease and [ CH₃OH ] will increase. A new equilibrium is established.
ii) Addition of CH₃OH:
The equilibrium will shift towards left, so more of the Hydroxide will dissociate. Concentrations of H2 and CO will increase.
iii) Removal of CO
So system will try to increase [ CO ]. This is possible only by producing more CO from reverse reaction. So Hydroxide will dissociate further. So H2 and CO are produced more. Reaction shifts to the left side.
2 (Mn O4)⁻ (aq) + 5 SO₂ (g) ⇒ 2 Mn²⁺ (aq) + 5 (HSO4)⁻ (aq) + e⁻
=> Acidic medium: So Add H⁺ ions and H₂O available in the medium.
2 MnO₄⁻ + 5 SO₂ + 2 H₂O + H⁺ => 2 Mn²⁺ + 5 HSO₄⁻
The balancing process is longer with intermediate half reactions and intermediate products. We can use some coefficients a , b, c and then balance each element on both sides. It is more like algebra.
Alternatively, we write the half reactions for oxidation and reduction of the important elements involved. Then we add them together while balancing them. Further we remove the common compounds or ions or charges on both sides. We are left with the balanced final equation.
===========
a) Le Chatelier principle:
For a chemical reaction in a chemical equilibrium, if any change occurs in pressure, temperature or concentration then the system will favour the equilibrium towards the side that opposes the change.
The effect will be to shift the equilibrium towards the side that minimizes (absorbs) the changes. A new equilibrium is established. Kc or Kp will remain constant at the given temperature.
b) Application of Le Chatelier's principle
2 H₂ (g) + CO (g) ⇔ CH₃ OH (g)
Kc = [ CH₃ OH ] / { [ H₂ ]² * [ CO ] } = constant at a Temperature
i) H2 increased.
After the equilibrium is established, [ H₂ ] increases while others are constant, then equilibrium shifts to the right. So [ H₂ ] will decrease and [ CH₃OH ] will increase. A new equilibrium is established.
ii) Addition of CH₃OH:
The equilibrium will shift towards left, so more of the Hydroxide will dissociate. Concentrations of H2 and CO will increase.
iii) Removal of CO
So system will try to increase [ CO ]. This is possible only by producing more CO from reverse reaction. So Hydroxide will dissociate further. So H2 and CO are produced more. Reaction shifts to the left side.
:-)
hope the detailed explanation is good enough.
Thanks sir for the splendid answer.
Thank you so much sir!
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