Chemistry, 29.07.2019 11:00 kierafisher05

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or increasing the volume of the reaction vessel. complete the sentences to determine if the reaction is endothermic or exothermic and to determine which side of the equation has more molecules.

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Hello!

First, we can model the chemical reaction in the following way:

nA(g) ⇄ mB(g)

Now, we can start answering the two questions:

1) Determine if the reaction is endothermic or exothermic:

The reaction is endothermic. We know that because the fraction of products is increased when the temperature is increased. According to Le Chatelier's principle, the system will move in the opposing direction of the changes to establish a new equilibrium. We can model an endothermic reaction in the following way, knowing that it needs heat to be completed

nA(g) + Δ ⇄ mB(g)

So, adding heat in the left side of the reaction will displace the equilibrium in the opposing direction (the right side) increasing the fraction of products.

2) Determine which side of the equation has more molecules:

The right side of the equation has more molecules than the left side. According to Le Chatelier's Principle, in a gas phase reaction, increasing the volume results in a decrease in the internal pressure of the system, so the reaction shifts to the side with more molecules to restore the pressure of the system. As the fraction of products increases with the increase in volume, the right side will have more molecules. We can then express this reaction in the following way:

nA(g) ⇄ mB(g) n<m

Have a nice day!

Answer from: Quest

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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