What is enthalpy of combustion (per mole) of C3H8(g)

2 C2H2 + 5 02  >  4 CO2 + 2 H2O

Products  - Reactants ( all units are kJ/mo1):

(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1

-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1

-1255.6 kJ/mo1

-2044.0 kJ/mol

Explanation:

Answer on Edg 2020

Enthalpy of combustion (per mole) of [tex]C_4H_{10} (g)[/tex] is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

[tex]2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})][/tex]

We are given:

[tex]\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ[/tex]

Enthalpy of combustion (per mole) of [tex]C_4H_{10} (g)[/tex] is -2657.5 kJ

The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.

Explanation:

[tex]\Delta H_{f,CO_2}=-393.5 kJ/mol[/tex]

[tex]\Delta H_{f,H_2O}=-241.82 kJ/mol[/tex]

[tex]\Delta H_{f,C_4H_{10}}=-125.6 kJ/mol[/tex]

[tex]\Delta H_{f,O_2}=0 kJ/mol[/tex]

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

Enthalpy of Combustion of 2 moles of butane :

[tex]=\sum(\Delta H_f\text{of products})-\sum(\Delta H_f\text{of reactants})[/tex]

[tex]\Delta H_c=(8\Delta H_{f,CO_2}+10\Delta H_{f,H_2O})-(2\Delta H_{f,C_4H_{10}}-13\Delta H_{f,O_2})[/tex]

[tex]=(8 mol\times -393.5 kJ/mol+10 mol\times (-241.82 kJ/mol))-(2 mol\times (-125.6 kJ/mol)+13 mol\times 0 kJ/mol)=-5315 kJ[/tex]

Enthalpy of Combustion of 1 moles of butane :

[tex]\Delta H_c=\frac{5315 kJ}{2mol}=-2657.5 kJ/mol[/tex]

The Enthalpy of combustion of 1 mol of butane is -2657.5 kJ/mol.

-1255.6 kJ/mol

Explanation:

-2,044.0 kJ/mol

Explanation:

I just did the test on edg

A) –2,044.0

Explanation

Its right on edge

A. –2,657.5 kJ/mol

Explanation:

This is correct on ed-genuity, hope this helps! :)

B. –1255.6 kJ/mol

Explanation:

this is correct on ed-genuity, hope this helps! :)

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]

Explanation:

Combustion is a rapid oxidation chemical process that is accompanied by low energy shedding in the form of heat and light. Oxygen is the essential element for oxidation to occur and is known as a oxidizer. The material that oxidizes and burns is the fuel, and is generally a hydrocarbon, as in this case butane C4H10 (g)

The balanced reaction is:

2 C4H10   +      13 O2       →      8 CO2   +     10H2O

Note that a balanced equation must have the same amount of each atom in the reagents and in the products, as in the previous reaction.

The heat of formation is the increase in enthalpy that occurs in the formation reaction of one mole of a certain compound from the elements in the normal physical state (under standard conditions: at 1 atmosphere of pressure and at 25 degrees of temperature).

In literature you can obtain the following heats of formation of each of the molecules involved in the reaction:

Heat of formation of C4H10 = -125.7 kJ/mol

Heat of formation of water = -241.82 kJ/mol

Heat of formation of CO2 = -393.5 kJ/mol

For the formation of one mole of a pure element the heat of formation is 0, in this case we have as a pure compound the oxygen O2

You want to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction. For that you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (quantity of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

                                             = (-393.5X8) + (-241.82X10) - (-125.7X2)

                                             = -5314.8 kJ/mol

But, if you observe the previous balanced reaction, you can see that 2 moles of butane are necessary in combustion. And the calculation of the heat of reaction previously carried out is based on this reaction. This ultimately means that the energy that would result in the combustion of 2 moles of butane is -5314.8 kJ/mol.

Then, applying a rule of three can calculate energy required for the combustion of one mole of butane: if for the combustion of two moles of butane an enthalpy of -5314.8 kJ / mol is required, how much energy is required for the combustion of one mole of butane?

Enthalpy of combustion (for 1 mol of butane)=[tex]\frac{-5314.8 \frac{kJ}{mol} }{2}[/tex]

Enthalpy of combustion (for 1 mol of butane)=-2657.4 [tex]\frac{kJ}{mol}[/tex]