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navik [9.2K]
3 years ago
8

Predict whether the equilibria I) CH4(g) + H2O(g) ⇀↽ CO(g) + 3 H2(g), ∆H◦ = +206 kJ II) 2 SO2(g) + O2 ⇀↽ 2 SO3(g), ∆H◦ = −198 kJ

will shift toward products or reactants with a temperature increase. 1. I shifts toward products and II shifts toward reactants. 2. Unable to determine 3. Both I and II shift toward products.
Chemistry
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

(1) I shifts toward product and II shifts toward reactant.

Explanation:

Increasing the temperature of an endothermic reaction (∆H is positive) shifts the equilibrium position to the right thus favoring product formation.

Increasing the temperature of an exothermic reaction (∆H is negative) shifts the equilibrium position to the left thus favoring the backward reaction.

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The substance that is doing the dissolving is referred to as?
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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
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Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

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4 years ago
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