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stira [4]
2 years ago
11

Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the sm

allest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.
Chemistry
1 answer:
Firdavs [7]2 years ago
7 0

Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

nFE^o_{cell}=RT\ln K

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

  • Higher E^o_{cell} higher will be the value of K.
  • Lower E^o_{cell} lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; E^o_{cell} =-0.030 V

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Explanation : Given,

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Formula used :

\text{Molarity}=\frac{\text{Moles of }KNO_3}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{0.500mol}{0.500L}=1.00mole/L=1.00M

Therefore, the molarity of solution is, 1.00 M

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How many grams are in 2.3 x10-4 moles of calcium phosphate ca3(po3)2
Airida [17]

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=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution

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