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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: N3 H12 P O3

Explanation:

From the question :

N = 31.57% H = 9.10% P = 23.27%

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Divide each of the element by their respective relative atomic masses.

N = 31.57 / 14 = 2.26

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The boiling point elevation of an aqueous sucrose solution is found to be 0.39°C.
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Answer:

130.4 grams of sucrose, would be needed to dissolve in 500 g of water.

Explanation:

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