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lara31 [8.8K]
3 years ago
14

How does Gibbs energy, enthalpy, and entropy determine whether a process will be a spontaneous or nonspontaneous

Chemistry
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

Because, Gibbs eats more of balanced diet, and that's why the process should be carried out.

Explanation:

I can say Gibbs is healthy and eats balanced diet so Gibbs energy, enthalpy, and entropy should determine whether a process will be a spontaneous or no spontaneous

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Onsider the following reaction at equilibrium:
11111nata11111 [884]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. H_2O is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. H_2 is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

5 0
4 years ago
What all indicates a CHEMICAL change? *
myrzilka [38]

one substance becomes two new substances

6 0
4 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
3 years ago
Please help i will give extra points
GREYUIT [131]

Answer:

B. Coal

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3 0
3 years ago
The iron content of foods can be determined by dissolving them in acid (forming Fe3+), reducing the iron(III) to iron(II), and t
____ [38]

Answer:

  • <em>Oxidation half-reaction</em>:

Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

  • <em>Reduction half-reaction</em>:

Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

Explanation:

The reaction that takes place is:

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The <em>oxidation half-reaction</em> is:

  • Fe²⁺(aq) →  Fe³⁺(aq) + 1e⁻

It is an oxidation because the oxidation state of Fe increases from 2+ to 3+.

The <em>reduction half-reaction</em> is:

  • Ce⁴⁺(aq) + 1e⁻ →  Ce³⁺(aq)

It is a reduction because the oxidation state of Ce decreases from 4+ to 3+.

4 0
3 years ago
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