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faust18 [17]
2 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]2 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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goldenfox [79]
Hello! Let me try to answer this :)

Thanks and please correct if there are any mistakes ^ ^

6 0
3 years ago
A drop of water (H20) has a mass of 2.34 x 10e-2 grams. How many
creativ13 [48]

Answer:

2333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

Explanation:

2x3=1000

3 0
2 years ago
If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
Zigmanuir [339]
<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0
3 years ago
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Elanso [62]
A change in position over a certain amount of time
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3 years ago
2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
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Answer 12.2g is the answer

4 0
2 years ago
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