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inessss [21]
3 years ago
10

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a

gas and can therefore be separated from solid impurities: Ni(s) + 4CO(g) ⇌ Ni(CO)4(g) Given that the standard free energies of formation of CO(g) and Ni(CO)4(g) are −137.3 and −587.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 58.0°C. Assume that ΔG o f is temperature-independent.
Chemistry
1 answer:
igor_vitrenko [27]3 years ago
4 0

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

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Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but
alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

<em>Where A is neutral base and AH⁺ is protonated form</em>

The basic dissociation constant, kb, is:

K_{b} = \frac{[AH^+][OH^-]}{[A]}

As pH in stomach is 2,5:

[OH] =10^{-[14-pH]}

[OH] = 3,16x10⁻¹² M

Thus:

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]} <em>(1)</em>

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

221359 = \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

0,0127 = \frac{[AH^+]}{[A]}

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

314456 = \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}

347851= \frac{[AH^+]}{[A]}

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

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3 years ago
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3 years ago
The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas
Contact [7]

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

PV=n_1RT

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

n_1 = number of moles of methane gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

(9.61atm)\times (3.06L)=n_1\times RT

n_1=\frac{29.4}{RT}

Now we have to calculate the moles of oxygen gas.

PV=n_2RT

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

n_2 = number of moles of oxygen gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

(1.75atm)\times (6.65L)=n_2\times RT

n_2=\frac{11.6}{RT}

Now we have to determine the final pressure in the system after mixing the gases.

P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}

where,

P_{total} = final pressure of gas = ?

V_{total} = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

Now put all the given values in the ideal gas equation, we get:

P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}

P_{total}=4.22atm

Therefore, the final pressure in the system is 4.22 atm.

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3 years ago
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3 years ago
The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
AlladinOne [14]

Answer:

64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

7 0
4 years ago
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