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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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I can help you with part 1
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3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

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3 years ago
Why is a cis-1,3-disubstituted cyclohexane more stable thanits<br> trans isomer?
anzhelika [568]

Answer:

Explanation has been given below

Explanation:

  • In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is present
  • In two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.
  • Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option.
  • Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.
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Why can't we separate azeotropes with the distillation process?.
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Fynjy0 [20]

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Now

\\ \rm\Rrightarrow \chi_{C_6H_6}

\\ \rm\Rrightarrow \dfrac{\chi_{C_6H_6}}{\chi_{sol^n}}

\\ \rm\Rrightarrow \dfrac{0.8}{0.8+0.2}

\\ \rm\Rrightarrow \dfrac{0.8}{1}

\\ \rm\Rrightarrow 0.8

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