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1 month ago

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A sample of chlorine gas is held at a pressure of 1023.6 Pa. When the pressure is decreased to 811.4 Pa the volume is 25.6 L. Wh

at was the original volume?

1 answer:

eduard [2.7K]1 month ago

6 0

Response:

V1 = 20.3L

Clarification:

P2 = 811.4Pa

V2 = 25.6L

P1 = 1023.6Pa

V1 =?

To answer this query, we will utilize Boyle's law, which states that the volume of a gas at constant temperature is inversely related to its pressure.

In mathematical terms,

V = k / P, where k = PV

The relationship can be defined as P1 × V1 = P2 × V2 = P3 × V3 =......=Pn × Vn

This simplifies to P1 × V1 = P2 × V2

Let’s rearrange for V1

V1 = (P2 × V2) / P1

Substituting values gives

V1 = (811.4 × 25.6) / 1023.6Pa

So, V1 = 20771.84 / 1023.6

This results in V1 = 20.29L, rounded to 20.3L

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Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.

This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.

The balanced equation for $CH_4$ appears as follows,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced reactions are outlined as follows,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

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(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

Therefore, the expression for the enthalpy of the reaction is,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Hence, the enthalpy change for this reaction is, 201.9 kJ

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