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3 years ago

PLEASE HELP !!

Chemistry

1 answer:

AVprozaik [17]3 years ago

3 0

Noble gas located in the region D

<h3>Further explanation</h3>

Noble gas are located in group 18 in the periodic table consisting of helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), have a stable electron configuration where the valence electrons are 8(full outer shell of electrons), so it is very hard to react because of its stability

If we look at the periodic system of the elements, the noble gases are on the far right, so that in Figure 8 it is located in area D

And the other region :

A : alkali elements(group 1),

B : alkaline earth elements(group 2)

C : halogen elements(group 17)

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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

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

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

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

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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:

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

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

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