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3 years ago

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Fe(NO3)2 not sure how to get the oxidation numbers of all elements

1 answer:

Shtirlitz [24]3 years ago

4 0

You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!

group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1

we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!

Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.

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Aleks [24]

Answer:

$K_{p}=4.35\times 10^{-4}$

Explanation:

We know, $K_{p}=K_{c}(RT)^{\Delta n}$

where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and $\Delta n$ is difference in sum of stoichiometric coefficient of products and reactants

Here $\Delta n=(2)-(2+1)=-1$ and T = 311 K

So, $K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}$

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3 years ago

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit

kati45 [8]

<u>Answer:</u> The $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

$CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

The expression of $K_p$ for the given reaction:

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

We are given:

$p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm$

Putting values in above equation, we get:

$K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = $5.34\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol$

Hence, the $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

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