Chemistry, 04.02.2020 22:56 stephen4438

Compare how entropy changes for the following two systems:

system a: two gases mix when the valve separating two containers is opened.

system b: a solid powder decomposes to form a solid product and a gas product.

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Answer from: tristenmathews

Explanation:

Entropy is defined as the amount of randomness present within the molecules of a substance.

So, when in system A, two gases mix together then it means there will be increase in number of molecules. As a result, there will be increase in number of collisions which will lead to increase in randomness of molecules.In system B, a solid powder is decomposing to given a solid product and a gas.

As gas molecules are held by weak forces so, they will rapidly move from one place to another. Hence, there will be increase in randomness of the molecules.

As a result, there will also be increase in entropy of system B.

A. Entropy increases

B. entropy increases

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

A. Two gases mix when the valve separating two containers is opened: The gases will intermix and thus there will be more disorder in the system.

B. A solid powder decomposes to form a solid product and a gas product.

The energy is absorbed by the solid reactant and thus it is endothermic and the solid reactants are converting to gaseous products, thus the entropy increases.

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Answer from: Quest

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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 :

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

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Compare how entropy changes for the following two systems:

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