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3 years ago

Какая масса хлора и какое количество вещества необходимо для реакции с 12г магния при получении хлорида магния?

1 answer:

7 0

Mg + Cl₂ = MgCl₂M(Mg) = 24г/моль m 1 моль Mg = 24 г.По условию задачи дано 12г. Mg Количество вещества n(Mg) =12÷24=0,5 мольРассуждаем: по уравнению реакции с 1 моль магния реагирует 1 моль хлора, следовательно с 0,5 моль будет реагировать 0,5 моль хлора.1 моль хлора при н.у. занимает объем 22,4л. , тогда 0,5 моль хлора займет:0,5х22,4л.= 11,2л. Ответ: Для взаимодействии 12 г. магния потребуется 11,2 л. хлора.

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A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

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A. The equilibrium constant K is defined as

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In any case

aA +Bb equilibrium Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

$ln(\frac{K_{1}}{K_{2}})=\frac{-delta H^{o}}{R}*(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

Where: H is reaction enthalpy, R is the gas constant and T temperature.

Clearing the equation for $K_{2}$ we get:

$K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

Here we can study two cases: when delta $H^{o}$ is positive (exothermic reactions) and when is negative (endothermic reactions)

For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so $K_{2}$ is greater that $K_{1}$ and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so $K_{2}$ will be less than $K_{1}$ the forward reactions is not favored.

${e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}$

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