# 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

1)

Enthalpy of combustion.

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.

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.

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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