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2 years ago

Estimate ΔG°rxn for the following reaction at 387 K. HCN (g) + 2 H2 (g) → CH3NH2 (g) ΔH° = −158.0 kJ; ΔS° = −219.9

Chemistry

1 answer:

Lina20 [59]2 years ago

4 0

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

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Answer:

Based on the information, the compound is a phospholipid.

Explanation:

Phospholipids are made up of a fatty acid tail which is hydrophobic in nature and a head which comprises of phosphates that is hydrophilic in nature. Hence, phospholipids are amphiphilic compounds so they will be partially soluble in water and will allow water-soluble substances to mix with fats.

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2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

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500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = <em>190,1L</em>

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

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I hope it helps!

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2 years ago

If a zero order reaction has a rate constant k of 0.0416Mmin and an initial concentration of 2.29 M, what will be its concentrat

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Answer:

The concentration after 20 mins is 0.832 M

Explanation:

Zero order rate law is given by;

R = K [A₀]⁰

A zero order reaction, rate is independent of the initial concentration

R = K

Where;

R is the rate of reaction

K is the rate constant = 0.0416 M/min

Since R = K,

Then, R = 0.0416 M/min

After 20 min, the concentration will be;

A = Rt

A = (0.0416 M/min)(20 min)

A = 0.832 M

Therefore, the concentration after 20 mins is 0.832 M

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2 years ago

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