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3 years ago

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For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium co

nstant at a certain temperature is 2.70 . 2.70. At this temperature, calculate the number of moles of NO 2 ( g ) NO2(g) that must be added to 2.99 mol SO 2 ( g ) 2.99 mol SO2(g) in order to form 1.30 mol SO 3 ( g ) 1.30 mol SO3(g) at equilibrium.

1 answer:

valina [46]3 years ago

3 0

Answer : The number of moles of $NO_2$ added must be, 0.37 mol

Explanation :

The given chemical reaction is:

$SO_2(g)+NO_2(g)\rightleftharpoons SO_3(g)+NO(g)$

Initial mol. 2.99 x 0 0

At eqm. (2.99-1.30) (x-1.30) 1.30 1.30

= 1.69

The expression for equilibrium constant is:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

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

$2.70=\frac{(1.30)\times (1.30)}{(1.69)\times (x-1.30)}$

x = 1.67 mol

The moles of $NO_2$ added = (x-1.30) = (1.67-1.30) = 0.37 mol

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Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

$G=G(T,P,n_1,n_2)$

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Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable $(T,P,n_j)$ are kept constant are known as partial molar Gibbs free energy of $i^{th}$ component.

For a substance in a mixture, the chemical potential $(\mu)$ is defined as the partial molar Gibbs free energy.

The expression will be:

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3 years ago

Explain why the nucleus of an atom is positively charged but the overall charge of an atom is neutral

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The nucleus is positively charged because it contains all of the protons however the charge of an atom is neutral because there is an equal amount of protons (positive) and electrons (negative) meaning the opposite charges cancel each other out therefore making the atoms overall charge neutral.

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2 years ago

g On the basis of their respective pKa values, determine which is the stronger acid, acetic acid (pKa 5 4.7) or benzoic acid (ma

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Answer:

Benzoic acid is the stronger acid

Explanation:

Weak acids do not dissociate completely in the solution. They exists in equilibrium with their respective ions in the solution.

The extent of dissociation of the acid furnising hydrogen ions can be determined by using dissociation constant of acid ( $K_a$ ).

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$HA \rightleftharpoons A^- + H^+$

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$pK_a$ is defined as the negative logarithm of $K_a$ .

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All can be summed up as:

The less the value of $pK_a$ , the more the $K_a$ is and the more the acid dissociates and the more the stronger is the acid.

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$pK_a$ of acetic acid = 54.7

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3 years ago

Diamond is forever" is one of the most successful advertising slogans of all time. but is it true? for the reaction shown below,

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The given reaction is

$C _{(Diamond)}\rightarrow C_{(Graphite)}$

An element can exist in 2 or more different forms which have totally different chemical and physical properties. They are known an allotropes.

Diamond and Graphite are allotropic forms of carbon. In the given reaction, diamond is changing to graphite and we have to find out the standard free energy of this reaction. This will help us to find out whether this reaction is spontaneous at 298 K or not.

The following data is needed for calculations which is taken from standard reference table.

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Step 1: Find ΔH⁰ rxn for the given reaction.

The formula to calculate ΔH⁰ rxn is given below.

$\bigtriangleup H^{0}_{rxn}= H_{f}(product)- H_{f}( reactant)$

We have graphite on product side and diamond on reactant side.

Therefore, $\bigtriangleup H^{0}_{rxn}= H_{f}(graphite)- H_{f}(diamond)$

Let us plug in the values given above.

$\bigtriangleup H^{0}_{rxn}= 0 - 1.895 kJ/mol$

$\bigtriangleup H^{0}_{rxn}= - 1.895 kJ/mol$

ΔH⁰ rxn for the given reaction is -1.895 kJ/mol

Step 2 : Find ΔS⁰ rxn for the given reaction.

The formula to calculate ΔS⁰ rxn is

$\bigtriangleup S^{0}_{rxn}= S^{0}(product)- S^{0}( reactant)$

$\bigtriangleup S^{0}_{rxn}= S^{0}(graphite)- S^{0}(diamond)$

$\bigtriangleup S^{0}_{rxn}= (5.740J/mol.K) - (2.337 J/mol.K)$

$\bigtriangleup S^{0}_{rxn}= 3.403 J/mol.K$

Let us convert this to kJ.

$\frac{3.403J}{mol.K}\times \frac{1 kJ}{1000J}= 3.403\times 10^{-3}kJ/mol.K$

ΔS⁰ rxn for the given reaction is 3.403 x 10⁻³ kJ/mol.K

Step 3: Find standard free energy ΔG⁰ rxn.

ΔG⁰ rxn for the given reaction is calculated as

$\bigtriangleup G^{0}_{rxn}= \bigtriangleup H^{0}_{rxn}- T\times \bigtriangleup S^{0}_{rxn}$

We have T = 298 K. Let us plug in the calculated values of ΔH⁰ rxn and ΔS⁰ rxn.

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [(298K)\times 3.403\times 10^{-3}kJ/mol.K]$

$\bigtriangleup G^{0}_{rxn}= - 1.895 kJ/mol - [1.014 kJ/mol]$

$\bigtriangleup G^{0}_{rxn}= - 2.909 kJ/mol$

The standard free energy change for the given reaction is -2.909 kJ/mol

The negative value of delta G⁰ suggests that the given reaction is spontaneous at room temperature. That means diamond will slowly convert to graphite. The speed of this reaction is extremely slow, but yet the reaction is taking place. So over a period of time diamond will become graphite.

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