Answer:
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Explanation:
Answer:
a. increase the temperature: the reaction will be shifted to the left side.
b. increase the pressure: the reaction will be shifted to the right side.
c. increase [SO₂]: so, the reaction will be shifted to the right side.
d. add a catalyst: it has no effect.
Explanation:
<em>Le Châtelier's principle</em><em> states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
a. increase the temperature:
∵ ΔH has a negative value, the reaction is exothermic.
- The heat can be represented as a part of the products.
- Increase the T will increase the concentration of the products "heat".
- So, the reaction will be shifted to the left side to suppress the increase in the concentration of products by increasing T.
<em>So, the reaction will be shifted to the left side.</em>
<em></em>
b. increase the pressure:
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 3.0 moles of gases and the products side (right) has 2.0 moles of gases.
- Thus, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).
<em>So, the reaction will be shifted to the right side.</em>
<em></em>
c. increase [SO₂]:
- Increasing [SO₂] will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of SO₂.
<em>So, the reaction will be shifted to the right side.</em>
<em></em>
d. add a catalyst:
- Catalyst increases the rate of the reaction without affecting the equilibrium position.
- Catalyst increases the rate via lowering the activation energy of the reaction.
- This can occur via passing the reaction in alternative pathway (changing the mechanism).
- The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
- in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
- with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
<em>So, it has no effect.</em>
<em></em>
Assuming that we are given the values of Ka and Kb, these can be rewritten and represented by the concentration of the Hydronium ion and Hydroxide Ion:
Hydronium ion: [H3O+]
Hydroxide ion: [OH-]
Ka is the equilibrium constant of an acid which dissociates to a hydronium ion and its cation.
Kb is the equilibrium constant of a base which dissociates to a hydroxide ion and its anion.
So,
Ka x Kb = [H3O+][OH-]
Answer:
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced molecular equation for the reaction of aqueous sodium sulfate, and aqueous barium nitrate is :
<span>Answer:
2COâ‚‚ (g) + 6Hâ‚‚ (g) → CHâ‚OCHâ‚ (g) + 3Hâ‚‚O (â„“)
Now make this reaction up from what you are given:
2 Ă— (COâ‚‚ (g) + 3Hâ‚‚ (g) → CHâ‚OH (l) + Hâ‚‚O (â„“).. â†H° = –131 kJ)
... (2CHâ‚OH (l) → CHâ‚OCHâ‚ (g) + Hâ‚‚O (â„“) .. â†H° = 8kJ)
––––––––––––––––––––––––––––––––––––––... add
2COâ‚‚ (g) + 6Hâ‚‚ (g) → CHâ‚OCHâ‚ (g) + 3Hâ‚‚O (â„“) . - exactly what you want!
â†H° = 2(–131 kJ) + 8 kJ = –254 kJ</span>