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3 years ago

Please help me understand how to...

Chemistry

1 answer:

KonstantinChe [14]3 years ago

3 0

Answer:

$m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Explanation:

Hello!

In this case, according to the reaction:

$4Cr+3O_2\rightarrow 2Cr_2O_3$

We can see there is a 4:2 mole ratio between chromium and chromium (III) oxide, this, for the given 56.2 g of chromium, the theoretical yield of the oxide product is computed down below:

$m_{Cr_2O_3}^{theoretical}=56.2gCr*\frac{1molCr}{52.0gCr}*\frac{2molCr_2O_3}{4molCr} *\frac{151.99gCr_2O_3}{1molCr_2O_3} =82.13gCr_2O_3$

Now, considering the 76.0-% yield for this reaction, the actual yield turns out:

$m_{Cr_2O_3}^{actual}=82.13gCr_2O_3*\frac{76.0gCr_2O_3}{100gCr_2O_3} \\\\m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Best regards!

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This application demonstrates how an understanding of free energy can explain seemingly unfavorable reactions proceeding, seemin

azamat

Answer: your question is incomplete, please let me assume this to be your question.

This application demonstrates how an understanding of free energy can explain seemingly unfavorable reactions proceeding, seemingly, spontaneously. Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are thermodynamically unfavorable (nonspontaneous), but proceed when they are coupled with other reactions.

ReactionA: P i + glucose ⟶ glucose-6-phosphate + H 2 O Δ G = 13.8 kJ / mol Reaction

ReactionB: P i + fructose-6-phosphate ⟶ fructose-1,6-bisphosphate + H 2 O Δ G = 16.3 kJ / mol

Reaction C: ATP + H 2 O ⟶ ADP + P i Δ G = − 30.5 kJ / mol

1. Which of these reactions is (are) unfavorable? Select all that apply.

2. Which of these reactions can be coupled so that overall reaction is favorable? Select all that apply.

3. What is the net change in free energy if one selection from part (b) is coupled so that the overall reaction is favorable?

THE ANSWERS ARE AS FOLLOWS

1. REACTION B IS UNFAVOURABLE

2. REACTION A AND C CAN BE COUPLED SO THAT OVERALL REACTION IS FAVOURABLE

3. Since reaction b is coupled the net change becomes

13.8+16.3= 30.1

Therefore the net free change becomes

30.5 - 30.1 = 0.4kj/mol

Explanation: The glycolysis pathway

is to describe the oxidation of glucose to pyruvate, with the generation of ATP and NADH. This pathway is also known as the EMBDEN-MEYERHOF PATHWAY.

Reaction B is unfavorable because it is the conversation of glucose into an unstable form, that can be readily cleaved into 3-carbon units. The unfavorable fructose-6-phosphate is quickly consumed to favour the forward reaction.

Reaction A and Reaction C makes the overall reaction to be favoured, since the unfavorable reaction is reaction B, which is an intermediate reaction.

In calculation of net free energy the negative sign which shows Exothermic reaction is multiplied by a negative sign.

The final energy minus the initial energy. The reaction B is added to the initial energy because it is an intermediate reaction, and we were told that one part of it is coupled to the reaction.

Hope this has helped you to solve your question.

5 0

4 years ago

At 298 K, Kc = 1.45 for the following reaction 2 BrCl (g) Br2(g) + Cl2(g) A reaction mixture was prepared with the following ini

tester [92]

Answer:

[BrCl]=0.02934M

[Br2]=[Cl2]=0.03533M

Explanation:

First, consider the Kc definition:

$1.5=K_{c}=\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}$

It is necessary to define a new variable, 'x', as the amount of moles of Br2 that are produced. If 'x' moles of Br2 are produced, the moles of the compounds will be calculated as;

$n_{Br_{2}}=n_{{Br_{2}}^{0}} +x\\n_{Cl_{2}}=n_{{Cl_{2}}^{0}} +x\\n_{BrCl}=n_{{BrCl}}^{0}} -2x\\$

Where the zero superscript means the initial moles.

Dividing the last equations by the volume, which is constant because the reaction does not change the total moles number (2 moles of BrCl produce 2 moles, one of Cl2 and another of Br2), we have the molarity equations for all species:

$M_{Br_{2}}=M_{{Br_{2}}^{0}} +x\\M_{Cl_{2}}=\\M_{BrCl}=M_{{BrCl}}^{0}} -2x\\$

And now 'x' is a change in molarity.

Replacing these in the Kc equation we have:

$1.45=\frac{({M_{{Br_{2}}}^{0}}+x)({M_{{Cl_{2}}}^{0}}+x)}{({M_{{BrCl}}^{0}}-2x)^{2}}$

Where the only unknown is 'x'. So, let's solve the equation:

$1.45=\frac{(0.03+x)(0.03+x)}{(0.04-2x)^{2} } \\1.45(0.04-2x)^2=(0.03+x)^2\\1.45(0.0016-0.16x+4x^2)=0.0009+0.06x+x^2\\4.8x^2-0.292x+0.00142=0\\x_{1}=0.0555\\x_{2}=0.00533$

The result $x_{1}=0.0555$ lacks of sense because it will give a negative concentration for BrCl, so the result is $x_{2}=0.00533$ <u>.</u>

Applying the result, the concentrations at equilibrium are:

$M_{Br_{2}}=M_{Cl_{2}}=0.03+0.00533=0.03533M\\M_{BrCl}=0.04-2*0.00533=0.02934M$

If you calculate Kc with this concentrations it will give 1.45 as a result.

Greets, I will be happy to solve any doubt you have.

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