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3 years ago

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Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2

HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? Δ G0f(kJ/mol) -110.9 87.6 51.3 -237.1

1 answer:

Pavlova-9 [17]3 years ago

7 0

Answer:

The equilibrium constant K = 1.15*10^-9

Explanation:

<u>Given:</u>

ΔG°f(HNO3) = -110.9 kj/mol

ΔG°f(NO) = 87.6 kj/mol

ΔG°f(NO2) = 51.3 kj/mol

ΔG°f(HNO3) = -237.1 kj/mol

<u>To determine:</u>

The equilibrium constant (K) for the given reaction

<u>Calculation:</u>

The chemical reaction is:

$2HNO3(aq) + NO(g) \rightarrow 3 NO2(g) + H2O(l)$

The equation that relates the standard free energy change ΔG° to the equilibrium constant K is:

$\Delta G^{0}= -RTlnK$

(or) $K = e^{-\Delta G^{0}/RT}----(1)$

where R = gas constant = 8.314 J/mol-K

T = temperature in Kelvin

$\Delta G^{0}=\sum n_{p}\Delta G^{0}f(products)-\sum n_{r}\Delta G^{0}f(reactants)$

where n(p) and n(r) are the number of moles of the products and reactants respectively

Therefore for the given reaction:

$\Delta G^{0}=[3\Delta G^{0}f(NO2)+3\Delta G^{0}f(H2O)]-[2\Delta G^{0}f(HNO3)+1\Delta G^{0}f(NO)]$

Substituting the given values for ΔG°f:

$\Delta G^{0}=[3\Delta G^{0}f(51.3)+3\Delta G^{0}f(-237.1)]-[2\Delta G^{0}f(-110.9)+1\Delta G^{0}f(87.6)]$

ΔG° = + 51 kJ

Substituting the calculated ΔG° in equation (1) at T = 298 K gives:

$K = e^{-\51000/8.314*298}=1.15*10^{-9}$

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Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

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<u>Answer:</u>

<u>(a):</u> The theoretical yield of silicon is 72.33 kg.

<u>(b):</u> The percent yield of the reaction is 91.25 %.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For </u> $SiO_2$ <u>:</u>

Given mass = 155.0 kg = 155000 g (Conversion factor: 1 kg = 1000 g)

Molar mass = 60 g/mol

Putting values in equation 1:

$\text{Moles of }SiO_2=\frac{155000g}{60g/mol}=2583.3mol$

<u>For carbon:</u>

Given mass = 78.2 kg = 78200 g

Molar mass = 12 g/mol

Putting values in equation 1:

$\text{Moles of carbon}=\frac{78200g}{12g/mol}=6516.67mol$

The chemical equation for the reaction of silicon dioxide and carbon follows:

$SiO_2+2C\rightarrow Si+2CO$

By stoichiometry of the reaction:

1 mole of $SiO_2$ reacts with 2 moles of carbon

So, 2583.3 moles of $SiO_2$ will react with = $\frac{2}{1}\times 2583.3=5166.4mol$ of carbon

As the given amount of carbon is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, $SiO_2$ is considered a limiting reagent because it limits the formation of the product.

<u>For (a):</u>

By stoichiometry of the reaction:

1 mole of $SiO_2$ produces 1 mole of silicon

So, 2583.3 moles of $SiO_2$ will produce = $\frac{1}{1}\times 2583.3=2583.3mol$ of silicon

Since the molar mass of silicon = 28 g/mol

Putting values in equation 1:

$\text{Mass of Si}=2583.3mol\times 28g/mol=72332.4g=72.33 kg$

Hence, the theoretical yield of silicon is 72.33 kg.

<u>For (b):</u>

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Measured value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Measured value of silicon = 66.0 kg

Theoretical value of silicon = 72.33 kg

Putting values in equation 1:

$\% \text{yield}=\frac{66.0kg}{72.33kg}\times 100\\\\\% \text{yield}=91.25 \%$

Hence, the percent yield of the reaction is 91.25 %.

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