Chemistry, 22.04.2020 03:35 riaahroo2378

If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.

Answers

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

$HN_3(g)\rightarrow N_2(g)+H_2(g)$

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

$2HN_3(g)\rightarrow 3N_2(g)+H_2(g)$

As we are given:

The pressure of pure HN_3 = 3.0 atm

$p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm$

From the reaction we conclude that:

Number of moles of N_2 = 3 mol

Number of moles of H_2 = 1 mol

Now we have to calculate the mole fraction of N_2 and H_2

$\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75$

and,

$\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25$

Now we have to calculate the partial pressure of N_2 and H_2

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

p_i = partial pressure of gas

p_T = total pressure of gas = 6.0 atm

X_i = mole fraction of gas

$p_{N_2}=X_{N_2}\times p_T$

$p_{N_2}=0.75\times 6.0atm=4.5atm$

and,

$p_{H_2}=X_{H_2}\times p_T$

$p_{H_2}=0.25\times 6.0atm=1.5atm$

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Answer from: Quest

it should be all of them because it asks if you included all of those in your answer

explanation:

Answer from: Quest

The answer is 65.3312

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If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction...