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2 years ago

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What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

3OH -->4 CH4 + O2 ∆H
=252.8 Kj

1 answer:

vodka [1.7K]2 years ago

5 0

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

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A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

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Answer:

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Explanation:

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A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

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2CH4(g)⟶C2H4(g)+2H2(g)

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Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of $CH_4$ will be,

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

The intermediate balanced chemical reaction will be,

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

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

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

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

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

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

The expression for enthalpy of the reaction will be,

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

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

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

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Answer:

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