Chemistry, 10.02.2020 21:03 cshopholic4921

A chemical equilibrium between gaseous reactants and products is shown.

N2(g) + 3H2(g) ⇌ 2NH3(g)

How will the reaction be affected if the pressure on the system is decreased? (3 points)

It will shift toward the reactant side as there is lower pressure on the reactant side.

It will shift toward the product side as there is higher pressure on the product side.

It will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

It will shift toward the product side as there are a fewer number of moles of gas on the product side.

Answers: 3

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Answers

Thrid statement: It will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

Explanation:

In the chemical equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), two molecules of N₂(g) react with 3 molecules of H₂(g) to form 2 molecules of NH₃(g).

Hence, in the reactant (left) side there are a total of 4 molecules reacting, while in the product (right) side there are a total of 2 molecules produced.

Following Le Chatelier's principle, an increase of the pressure on the system must be followed by a shift of the equilibrium to counteract that disturbance.

Hence, if the pressure on the system is decreased, to oppose this distrubance, the reaction will shift in the direction that increases the pressure, and that is to the side that more molecules are produced (because the pressure is proportional to the number of molecules).

As stated, the stoichiometry shows that there are more molecules on the left side. Then, you conclude that the reaction will shift toward the reactant side as there are a greater number of moles of gas on the reactant side.

Answer from: veikkoaval

just wanted to confirm that it is the third statement

Explanation:

just took the quiz :)

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

thats a little if u want the answer just look it up on google thats what i do haha : )

Draw the orbitals for the elements and give their number of electrons

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A chemical equilibrium between gaseous reactants and products is shown.

N2(g) + 3H2(g) ⇌...

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