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vesna_86 [32]
3 years ago
9

Calculate the standard potential, e∘, for this reaction from its equilibrium constant at 298 k. x(s)+y4+(aq)↽−−⇀x4+(aq)+y(s)k=4.

98×10−5
Chemistry
1 answer:
laila [671]3 years ago
6 0
First, we need to get the number of moles:

from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  

∴ the number of moles n = 4

we are going to use this formula:

㏑K = n *F *E/RT

when K is the equilibrium constant = 4.98 x 10^-5

and F is Faraday's constant = 96500

and the constant R = 8.314

and T is the temperature in Kelvin = 298 K

and n is number of moles of electrons = 4 

so, by substitution:

㏑4.98 x 10^-5 = 4*96500*E / 8.314*298

∴E = -0.064 V
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Answer:

The spontaneous reactions are:

B. C_{2}H_{4} + H_{2} \overset{Rh(I)}{\rightarrow}C_{2}H_{6} ; <u>ΔG= −150.97 kJ/mol  </u>

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Explanation:  

Gibbs free energy , denoted by ΔG, is the <u>quantitative measure of the favorability or spontaneity</u> of a given process or chemical reaction that is carried out at constant temperature (T) and pressure (P).

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Here, ΔH is the change in enthalpy and ΔS is the change in the entropy

For a given chemical reaction to be <u>spontaneous or favorable</u>, ΔG should be negative (ΔG < 0).

A. DHAP ⇌ glyceraldehyde-3-phosphate ; <u>ΔG = 3.8 kJ/mol  </u>

<u>The value of ΔG is positive (ΔG > 0). Therefore, this reaction is not spontaneous.</u>

B. C_{2}H_{4} + H_{2} \overset{Rh(I)}{\rightarrow}C_{2}H_{6} ; <u>ΔG= −150.97 kJ/mol  </u>

 <u>The value of ΔG is negative (ΔG < 0). Therefore, this reaction is spontaneous.</u>

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C. glutamate + NAD⁺ + H₂O ⟶ NH⁴⁺ +α-ketoglutarate + NADH + H⁺ ; <u>ΔG = 3.7 kcal/mol   </u>

<u>The value of ΔG is positive (ΔG > 0). Therefore, this reaction is not spontaneous.</u>

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D. C₆H₁₃O₉P + ATP ⟶ C₆H₁₄O₁₂P₂ + ADP; <u>ΔG = −14.2 kJ/mol  </u>

 <u>The value of ΔG is negative (ΔG < 0). Therefore, this reaction is spontaneous.</u>

<u />

E. L -malate + NAD⁺ ⟶ oxaloacetate + NADH + H⁺; <u>ΔG = 29.7 kJ/mol </u>

<u>The value of ΔG is positive (ΔG > 0). Therefore, this reaction is not spontaneous.</u>

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