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faust18 [17]
3 years ago
7

8. Consider the following reaction, for which ΔG° = −21.7 kJ/mol at 25°C: A2B(g) ↔ 2 A(g) + B(g) At the moment that 0.10 mol A2B

(g), 0.10 mol A(g), and 0.050 mol B(g) are placed into a 10.0 L flask at 25°C, which of the following would be true at that initial moment? a. Qc > Kc for this system at 25°C b. Qc < Kc for this system at 25°C c. ΔG < 0 d. ΔG >0 e. Qc = Kc for this system at 25°C f. rxn will shift to products g. rxn will shift to reactants (1) a, d, g (2) b, c, f (3) e only (4) b, d, g (5) a, c, f
Chemistry
1 answer:
Mrac [35]3 years ago
5 0

Answer:

(2) b, c, f

Explanation:

Hello,

In this case, for the undergoing chemical reaction:

A_2B(g) \rightleftharpoons 2 A(g) + B(g)

The Gibbs free energy of reaction at 25 °C is related with an equilibrium constant of:

K=exp(-\frac{\Delta G^0 }{RT} )=exp[-\frac{-21700J/mol}{8.314J/(mol*K)*298.15K}]=1.58x10^{-4}

Now, at the given moment, the reaction quotient turns out:

Q=\frac{(\frac{0.10mol}{10.0L})^2(\frac{0.050mol}{10.0L})}{(\frac{0.10mol}{10.0L})} =5x10^{-5}

Thus, since Q<K, the reaction will have too much reactants that will produce more products, shifting the reaction rightwards to them.

Hence, with the given information the statements b, c and f are correct, so the answer is: (2) b, c, f.

Best regards.

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Answer:

V₁  = 374.71  mL

Explanation:

Given data:

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Solution:

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Final temperature = 86°C (86+273 = 359 k)

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

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V₁/T₁ = V₂/T₂

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Answer:

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Explanation:

To answer the question given above,

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