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valentinak56 [21]
3 years ago
8

The activation energy for the reaction no2(g)+co(g)⟶no(g)+co2(g) is ea = 100 kj/mol and the change in enthalpy for the reaction

is δh = -250 kj/mol . what is the activation energy for the reverse reaction?
Chemistry
1 answer:
IrinaK [193]3 years ago
7 0
Answer: 350 kJ/mol

Explanation:

1) <span>δh = -250 kj/mol means that the products of the forward reaction are 250 kJ/mol lower in the potential enery diagram.

2) So, the reverse reaction must compensate the 250 kJ/mol and go upper as much as the activation energy of the forward reaction.

This is: Ea reverse = Ea forward + | </span><span>δh |

You can see this clearer on the attached graph of an exothermic reaction.

Ea reverse = 100 kJ / mol + | -250 kJ/mol | = 100 kJ/mol + 250 kJ/mol = 350 kJ/mol
</span>
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(a)   Formula that shows relation between K_{c} and K_{p} is as follows.

                 K_c = K_p \times (RT)^{-\Delta n}

Here, \Delta n = 1

Putting the given values into the above formula as follows.

        K_c = K_p \times (RT)^{-\Delta n}

                  = 1.20 \times (RT)^{-1}

                  = \frac{1.20}{0.0820 \times 1073}

                  = 0.01316

(b) As the given reaction equation is as follows.

               CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)

As there is only one gas so ,

                p[CO_{2}] = K_{p} = 1.20

Therefore, pressure of CO_{2} in the container is 1.20.

(c)   Now, expression for K_{c} for the given reaction equation is as follows.  

             K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}

                        = \frac{x \times x}{(a - x)}

                        = \frac{x^{2}}{(a - x)}[/tex]

where,    a = initial conc. of CaCO_{3}

                  = \frac{22.5}{100} \times 9.56

                  = 0.023 M

          0.0131 = \frac{x^{2}}{0.023 - x}

                  x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

       % of CaCO_{3} remained = (\frac{0.017}{0.023}) \times 100

                                  = 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

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