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tangare [24]
3 years ago
14

The following reaction is exothermic. C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g)C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g) Predict the effect (shi

ft right, shift left, or no effect) of increasing and decreasing the reaction temperature.
Chemistry
2 answers:
inna [77]3 years ago
7 0

Answer:

Increasing temperature = balance will shift to the left

Decreasing temperature = balance will shift to the right

Explanation:

Step 1: Data given

The increase or decrease in temperature can have an influence on the position of the equilibrium.

If the temperature is increased, the system will ensure that less heat is released. So the balance will shift to the left.

When the temperature drops, however, the system will produce more heat: the balance will shift to the right.

Step 2: The balanced equation

C6H12O6(s) + 6O2(g) ⇌ 6CO2(g) + 6H2O(g)

This is an endothermic reaction

Step 3: Increasing the temperature

If the temperature were increased, the heat content of the system would increase.

In exothermic reactions, increase in temperature decreases the K value. This means less products will be formed. The balance will shift to the left.

Step 4: Decreasing the temperature

If the temperature were decreased, the heat content of the system would increase.

In exothermic reactions, decrease in temperature increases the K value. This means more products will be formed, less reactants. The balance will shift to the right.

elena55 [62]3 years ago
4 0

Answer:

According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.

Explanation:

C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)

We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.

According to Le Chatelier's principle,

1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.

2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.

3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.

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3 years ago
Which of the following is not a conjugate acid-base pair?
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A. H3O+/OH−

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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
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Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

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