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Julli [10]
3 years ago
7

Help please !!!!!!!!

Chemistry
1 answer:
Elena L [17]3 years ago
5 0

Answer:

Option B. 2096.1 K

Explanation:

Data obtained from the question include the following:

Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =.?

Entropy is related to enthalphy and temperature by the following equation:

Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)

ΔS = ΔH / T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

ΔS = ΔH / T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K

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Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
Why do you have to have a certain number of reactants on one side?
Debora [2.8K]

Answer:

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Explanation:

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3 0
2 years ago
PLEASE HELPPPP!!!!!
Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3

Now put all the given values in this formula, we get the mass of nitric acid.

\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g

Therefore, the mass of nitric acid is, 214.234 grams.


6 0
3 years ago
Liquids do not have the energy to overcome the ___ and move to the gaseous state.
Ugo [173]

Answer - Inter-molecular attractions

Explanation-

As we know everything around us is made up of matter that means everything has molecules as their basic structure. The state of anything is decided by the spaces between the molecules.

The state of the objects that have strong inter-molecular attractions a solid and gradually the lesser will be in state of liquid and gas. The attraction between the molecules is overcome only when a certain amount of energy is provided from outside.  

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2 years ago
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Nastasia [14]

The sign of the entropy change, ΔS, for the following processes include:

  • I2 (s) + ½ Cl2 (g) ← → ICl (g) - positive
  • 3 Ag (s) + 4 HNO3 (aq) → 3AgNO3 (aq) + NO (g) + 2 H2O (l) - positive
  • Cl2 (g) → Cl2 (l) - negative
  • C5H12 (g) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (l) - negative

What is Entropy?

This is referred to the degree of randomness or disorderliness of a system and is denoted as S.

Entropy change is usually positive when solid or liquid reactant is converted into gas product and vice versa.

Read more about Entropy here brainly.com/question/419265

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6 0
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