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Novosadov [1.4K]
3 years ago
10

Explain the difference between a physical separation and a chemical separation

Chemistry
1 answer:
Gnom [1K]3 years ago
6 0

A chemical separation is a separation method that involves change in the chemical composition of one of the solutions that are to be separated in order to a new physical properties of one part reached leaving the rest alone.


For example, When we add a chemical to the solution that will precipitate only one part of the solution leaving the rest.


A physical separation is a separation in which solutions are separated on the basis of their physical properties.


For example, If we use physical separation method in separating two oils with different specific volumes then these two oils are separated on the basis of their settling.

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac
valentinak56 [21]

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

Now put all the given values in this formula, we get:

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

The expression for reaction quotient will be :

Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln Q

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

3 0
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Explanation:

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DerKrebs [107]

I believe it is D. Hope this helps you!

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