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cricket20 [7]
3 years ago
6

In the uncatalyzed reaction N2O4 (g) ⇌ 2 NO2 (g) the pressure of the gases at equilibrium are PN2O4 = 0.377 atm and PNO2 = 1.56

ATM AT 100ºc. What would happen to these pressures if a catalyst were added to the mixture?
Chemistry
1 answer:
larisa86 [58]3 years ago
8 0

Answer:

The pressures will remain at the same value.

Explanation:

A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.

While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.

Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).

The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.

Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.

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Answer:

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Explanation:

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Where E^{0}(reduction) and E^{0}(oxidation) represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation:         Mg-2e^{-}\rightarrow Mg^{2+} ;  E_{Mg^{2+}\mid Mg}^{0}=-2.38V

Reduction:         Fe^{3+}+3e^{-}\rightarrow Fe ; E_{Fe^{3+}\mid Fe}^{0}=-0.04V

So, E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V

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(2) Similarly as above, E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V

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(3) Similarly as above, E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V

Hence this pair will give non-spontaneous reaction.

(4)  Similarly as above, E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V

Hence this pair will give non-spontaneous reaction.

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