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3 years ago

Given the reaction: N2 + O2 = 2NO for which the Keq at 2273 K is 1.2 x 10-4

Chemistry

1 answer:

MAXImum [283]3 years ago

3 0

<u>Answer:</u>

(a): The expression of equilibrium constant is $K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$

(b): The equation to solve the concentration of NO is $[NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}$

(c): The concentration of NO is 0.0017 M.

<u>Explanation:</u>

The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term $K_{eq}$

(a):

The given chemical equation follows:

$N_2+O_2\rightarrow 2NO$

The expression for equilbrium constant will be:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$

(b):

The equation to solve the concentration of NO follows:

$[NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}$ ......(1)

(c):

Given values:

$K_{eq}=1.2\times 10^{-4}$

$[N_2]_{eq}=0.166M$

$[O_2]_{eq}=0.145M$

Plugging values in equation 1, we get:

$[NO]=\sqrt{(1.2\times 10^{-4})\times 0.166\times 0.145}$

$[NO]=\sqrt{2.88\times 10^{-6}}$

$[NO]=0.0017 M$

Hence, the concentration of NO is 0.0017 M.

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3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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