For Each Of The Following Reactions, Identify Another Quantity That Is Equal To Δh°Rxn.

For each of the following reactions, identify another quantity that is equal to delth degree rxn. 1. ch4(g) + 2o2(g) rightarrow co2(g) + 2h2o(i) enathalpy of combustion of ch4 enthaply of formation of co2(g) 4x bond energy of c – h – 4x bond energy of c – h 2. ch4(g) rightarrow c(g) + 4h(g) enthalpy of combustion of ch4 enthalpy of formation of c(g) 4x bond energy of c – h – 4x bond energy of c – h 3. c(graphite) + 2h2(g) rightarrow ch4(g) enthalpy of comusion of ch4 enthalpy of formation of ch4(g) 4x bond energy of c – h – 4x bond energy of c – h 4. c(g) + 4h(g) rightarrow ch4(g) enthalpy of comusion of c enthalpy of formation of ch4(g) 4x bond energy of c – h – 4x bond energy of c – h

Answers

1)

Enthalpy of combustion.

You are watching: For each of the following reactions, identify another quantity that is equal to δh°rxn.

2)

*

4x bond energy of C – H

3)

enthalpy of formation of CH₄(g)

4)

– 4x bond energy of C – H

Explanation:

The enthalpy of reaction is the heat change during the reaction.

It can be:

a) Enthalpy of combustion : the heat change during complete burning of one mole of a substance in presence of oxygen.

b) Enthalpy of formation: the heat change during formation of one mole of a substance from its constituent elements in their native state.

c) The difference in the bond energies of reactants and products.

*

For:

*

–<4XbondenergyofC-H>=-4XbondenergyofC-H” title=”3f9a2″>

Enthalpy of formation of ClO (ΔHf⁰) = 96.5 Kj/mole

Explanation:

Given that

chlorine-oxygen bond has an enthalpy of 243 kJ/mole , an oxygen-oxygen bond has an enthalpy of 498 kJ/mole.

*

(g) + (g) → ClO(g)

ΔHr = ∑ Bond energies

= x (Bond enthalpy of oxygen) – (Bond enthalpy of Cl – O bond)

= x 498 kj/mole – 243 kj/mole

= + 6 kj/mole

ClO(g) + (g) → ClO₂ (g)

ΔHr = Enthalpy of formation of ClO₂ – Enthalpy of formation of ClO

⇒Enthalpy of formation of ClO = Enthalpy of formation of ClO₂ – ΔHr

⇒Enthalpy of formation of ClO = 102.5 kj/mole – 6 kj/mole

⇒Enthalpy of formation of ClO = 96.5 kj/mole

The enthalpy of the formation of is coming out to be -65.3 kJ/mol

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as

*

The equation used to calculate enthalpy change is of a reaction is:

-sum ” title=”72c39″>

For the given chemical reaction:

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

*

–<(1times Delta H^o_f_{(SiO_2(s))})+(3times Delta H^o_f_{(C(s))})>” title=”6a7fe”>

We are given:

*

Putting values in above equation, we get:

*

–<(1times (-910.9))+(3times (0))>\Delta H^o_f_{(SiC(s))}=-65.3kJ/mol” title=”c8e18″>

Hence, the enthalpy of the formation of is coming out to be -65.3 kJ/mol.

Explanation:

CH 4 ( g ) + 2 O 2 ( g ) ⟶ CO 2 ( g ) + 2 H 2 O ( l )

enthalpy of formation of CO 2 ( g )

− 4 × bond energy of C H

4 × bond energy of C − H

enthalpy of combustion of CH 4

enthalpy of combustion of CH 4: This is because we usually calculate the enthalpy change of combustion from enthalpies of formation. This is given as;

ΔHoc = ∑ΔH∘f(products) − ∑ΔH∘f(reactants)

CH 4 ( g ) ⟶ C ( g ) + 4 H ( g )

1. enthalpy of combustion of CH4

2. enthalpy of formation of C(g)

3. 4× bond energy of C–H

4. –4× bond energy of C–H

4× bond energy of C–H: Bond breaking is an endothermic process. So it cannot be option 4, enthalpy of formation of a pure element is 0 KJ/mol. This is not a combustion reaction, so options 1 and 2 are aslo wrong.

c) C(graphite) + 2H2(g) CH4(g)

1. enthalpy of combustion of CH4

2. enthalpy of formation of CH4(g)

3. 4× bond energy of C–H

4. –4× bond energy of C–H

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. The correct option is; enthalpy of formation of CH4(g)

d) C(g) + 4H(g) CH4(g)

1. enthalpy of combustion of C

2. enthalpy of formation of CH4(g)

3. 4× bond energy of C–H

4. –4× bond energy of C–H

–4× bond energy of C–H; bond formation is always exothermic, it cannot be option 3. This is not a combustion reaction so it cannot be option 1. enthalpy of formation of a pure element is 0 KJ/mol, so it cannot be option 2.

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Explanation:

The chemical equation is as follows.

And, the given enthalpy is as follows.

*

;

*

= 102.5 kJ

Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

102.5 =

*

– <(2)(243)>” title=”c814b”>

102.5 =

*

102.5 – 12 =

*

x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

*

x = – 243″ title=”97422″>

= 339.5 – 243

= 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

The answer & explanation for this question is given in the attachment below.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

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A) the enthralling of formation for pure element in its standard state is always positivehope this : )
All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.

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