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At a fixed temperature, equal moles of N2 (g) and H2 (g) are mixed in a constant pressure container (the volume of the container changes in order to keep the pressure at a constant value). The N2 (g) and H2 (g) are allowed to react, producing NH3 (g): N2 (g) + 3 H2 (g) → 2 NH3 (g) If the initial volume of the container, before any reaction takes place, is 5.80 L, determine the volume of the container after the N2 (g) and H2 (g) have reacted to completion.

Respuesta :

Answer : The volume of the container after the nitrogen gas and hydrogen gas reacted to completion will be, 3.867 L

Explanation :

As we are given that the moles of [tex]N_2[/tex] and [tex]H_2[/tex] are equal. So, initial moles are:

[tex]n_{initial}=n+n=2n[/tex]

The balanced chemical reaction is,

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

When 'n' moles of [tex]H_2[/tex] react with [tex]\frac{n}{3}[/tex] moles of [tex]N_2[/tex] then it gives [tex]\frac{2n}{3}[/tex] moles of [tex]NH_3[/tex]

So, final moles are:

[tex]n_{final}=(n)-(\frac{n}{3})+(\frac{2n}{3})=\frac{4n}{3}[/tex]

According to the Avogadro's Law, the volume of the gas is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 5.80 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]n_1[/tex] = initial moles of gas = 2n mole

[tex]n_2[/tex] = final moles of gas = [tex]\frac{4n}{3}[/tex]

Now put all the given values in the above formula, we get the final volume of the gas.

[tex]\frac{5.80L}{V_2}=\frac{2n}{(\frac{4n}{3})}[/tex]

[tex]V_2=3.867L[/tex]

Therefore, the volume of the container after the nitrogen gas and hydrogen gas reacted to completion will be, 3.867 L

onyebuchinnaji

The final volume of the container after the nitrogen  [tex](N_2)[/tex]  and hydrogen  [tex](H_2)[/tex] have reacted completely is 3.87 L.

The given parameters;

  • initial volume of the container, V₁ = 5.8 L
  • [tex]N_2 \ + \ 3H_2 \ --> \ 2NH_3[/tex]

 

  • let the common number of moles of N₂ and H₂ = x

The initial number of moles of the gases before reaction = x + x = 2x

The number of moles the gases after reaction is calculated as follows;

[tex]N_2 \ + \ 3H_2 \ --> \ 2NH_3[/tex]

[tex]x \ \ \ \ \ \ \ \ \ \frac{x}{3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{2x}{3}[/tex]

[tex]n_{total} =n_f = x - \frac{x}{3} + \frac{2x}{3} = \frac{4x}3}[/tex]

Apply Avogadro's law, which states that the volume of a gas is directly proportional to its number of moles;

[tex]\frac{v_1}{v_2} = \frac{n_1}{n_2} \\\\v_2 = \frac{v_1n_2}{n_1} \\\\v_2 = \frac{5.8L \ \times\ \frac{4x}{3} }{2x} \\\\v_2 = \frac{5.8L \ \times \ 4x}{3 \ \times \ 2x} \\\\v_2 = 3.87 \ L[/tex]

Thus, the final volume of the container after the [tex]N_2[/tex] and [tex]H_2[/tex] have reacted completely is 3.87 L.

Learn more here:https://brainly.com/question/12678568

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