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ENTHALPY OF FORMATION, Δ_fH MCQ
IIT JEE

1) Standard molar enthalpy of formation of CO₂ is:

(IIT JEE 1997)

1) the standard molar enthalpy of combustion of gaseous carbon. 

2) the sum of standard molar enthalpies of formation of CO and O₂.

3) the standard molar enthalpy of combustion of graphite.

4) zero

Logic:

The standard molar enthalpy of formation of a compound, Δ_fH^o is defined as the amount of heat either liberated or absorbed when one mole of that compound is formed from its constituent elements in the standard state.

Also, the standard molar enthalpy of combustion of a substance is defined as the amount of heat liberated when one mole of that substance is burned completely in excess of air (so that complete oxidation is ensured).

Solution:

The thermochemical equation for formation of one mole of CO₂ under standard conditions can be written as:

C _(graphite) + O_2(gas) ----------> CO_2(gas) + heat

According to the definition, the heat liberated in above equation is nothing but the standard enthalpy of formation of CO₂.

However, above equation also represents combustion of graphite. Hence the amount of heat liberated can be considered as standard heat of combustion of graphite.

Note: 

i) The standard state of carbon is graphite since it is thermodynamically more stable than diamond. For O₂ and CO₂, gaseous state is the standard state.

ii) In combustion, carbon must be oxidized completely to its highest oxidation number. i.e., CO₂ must be the product and not CO. 

Conclusion:

The correct option is "3".

Related questions

2) The Δ_fH^o CO₂(g), CO(g), H₂O(l) are -393.5, -110.5, -241.8 kJ mol^-1 respectively. The standard enthalpy change (in kJ mol^-1) for the reaction CO₂(g) + H₂(g) -------> CO(g) + H₂O(l) is:

(IIT JEE - 2000) 

a) 41.2 

b) 524.1

c) -262.5

d) -41.2

Logic & Solution:

The thermochemical equations for the formations of CO₂(g), CO(g), H₂O(g) can be written as follows.

(1) C (graphite) + O₂(g)  --------> CO₂(g) ; Δ_fH₁ = -393.5 kJ mol^-1

(2) C (graphite) + 0.5O₂(g)  --------> CO(g) ; Δ_fH₂ = -110.5 kJ mol^-1

(3) H₂ (g) + 0.5O₂(g)  --------> H₂O(l) ; Δ_fH₃ = -241.8 kJ mol^-1

Now remove equation (1) from equations [(2) + (3)] to get the required equation:

CO₂(g) + H₂(g) -------> CO(g) + H₂O(l) 

Therefore, for above equation, standard enthalpy change,:

Δ_rH = ( Δ_fH₂ + Δ_fH₃) - Δ_fH₁ = [(-110.5) + (- 241.8)] - (-393.5)  = 41.2 kJ

Alternatively:

For any given reaction, 

Δ_rH = ∑Δ_fH^o_products - ∑Δ_fH^o_products

Therefore for the reaction:

CO₂(g) + H₂(g) -------> CO(g) + H₂O(l) 

Δ_rH = [Δ_fH^o(CO) + Δ_fH^o(H₂O)] - [Δ_fH^o(CO₂) + Δ_fH^o(H₂)] =  [(-110.5) + (- 241.8)] - [(-393.5) + 0] = 41.2 kJ

Note: Standard heat of formation of any element in its standard state is taken as zero.

Conclusion:

The correct option is: 'a'.

3) The species which by definition has zero standard molar enthalpy of formation at 298 K is:

(IIT JEE - 2010) 

a) CH₄(g) 

b) Br₂(g)

c) Cl₂(g)

d) H₂O(g)

Explanation:

As mentioned in above problem, the standard molar enthalpy of formation of element in its standard state is taken as zero. Since the standard state of Cl₂ is gas, its Δ_fH^o = 0.

However, Br₂ is also an element but its standard state is liquid.

Conclusion:

The correct option is: 'c'.

4) For which of the following equations is Δ_rH equal to Δ_fH?

(IIT JEE - 2003) 

a) CH₄(g) + Cl₂(g) -------> CH₂Cl₂(l) + 2HCl(g)

b) Xe(g) + 2F₂(g) --------> XeF₄(g)

c) 2CO(g) + O₂(g)  --------> 2CO₂(g) 

d) N₂(g) + O₃(g) ---------> N₂O₃(g)

Explanation:

Since Δ_fH is the amount of heat exchanged when one mole of a compound is synthesized from its constituent elements in their standard states, only for the equation given under the option 'b', the heat of reaction, Δ_rH equals to heat of formation, Δ_fH.

The equation given under option 'd' does not correspond to heat of formation of N₂O₃ since one of the reactants i.e., O₃ is not the standard state of oxygen element.

Conclusion:

The correct option is: 'b'.

Homework

1) What is standard state?