All categories

1 month ago

2CH4(g)⟶C2H4(g)+2H2(g)

Chemistry

1 answer:

Alekssandra [3K]1 month ago

5 0

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

Explanation:

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

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

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

You might be interested in

"solid potassium iodide decomposes into iodine gas and solid potassium. Write a a balanced chemical equation for this reaction"

Alekssandra [3086]

Now, construct a balanced equation:

$2KI (s) ----\ \textgreater \ 2K(s) + I_2 (g)$

$I_2$ exists in its gaseous form as a diatomic molecule.

3 0

1 month ago

During which time interval does the substance exist as both a liquid and a solid

lions [2927]

The Chemistry Regents is one of the four science Regents exams. The remaining three are Earth Science, Living Environment, and Physics. Passing at least one of these four exams is a requirement for high school graduation.

8 0

1 month ago

How to correctly solve this problem : 4.05Kg+567.95g+100.1g correct and best way

castortr0y [3046]

Refer to the attached document for the solution.

7 0

1 month ago

An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs

eduard [2782]

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

Explanation:

Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

6 0

16 days ago

what is the net ionic equation with its physical states? (NH4)2CO3(aq)+Ca(ClO4)2(aq)⟶CaCO3(s)+2NH4ClO4(aq)

lions [2927]

Answer: The net ionic equation is $CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)$

Explanation:

A double displacement reaction involves the exchange of ions. Chemicals that dissolve in water are marked with the symbol (aq), while those that do not dissolve and remain solid are shown with (s) after their formulas.

$(NH_4)_2CO_3(aq)+Ca(ClO_4)_2(aq)\rightarrow CaCO_3(s)+2NH_4ClO_4(aq)$

The ion-based representation of the equation is:

$2NH_4^+(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2ClO_4^{-}(aq)\rightarrow CaCO_3(s)+2NH_4^{+}(aq)+2ClO_4^-(aq)$

"Spectator ions" are the ions that do not participate in the chemical reaction, appearing on both sides of the equation in ionic form.

Ammonium and chlorate ions are present on both sides; thus, they do not factor into the net ionic equation.

Therefore, the net ionic equation is:

$CO_3^{2-}(aq)+Ca^{2+}(aq)\rightarrow CaCO_3(s)$

6 0

1 month ago