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# ENTHALPY OF FORMATION, ΔfH 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, ΔfHo 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) + O2(gas) ----------> CO2(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 ΔfHo 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) ; ΔfH1 = -393.5 kJ mol-1

(2) C (graphite) + 0.5O2(g)  --------> CO(g) ; ΔfH2 = -110.5 kJ mol-1

(3) H2 (g) + 0.5O2(g)  --------> H2O(l) ; ΔfH3 = -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,:

ΔrH = ( ΔfH2 + ΔfH3) - ΔfH1 = [(-110.5) + (- 241.8)] - (-393.5)  = 41.2 kJ

Alternatively:

For any given reaction,

ΔrH = ∑ΔfHoproducts - ∑ΔfHoproducts

Therefore for the reaction:

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

ΔrH = [ΔfHo(CO) + ΔfHo(H2O)] - [ΔfHo(CO2) + ΔfHo(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 ΔfHo = 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 ΔrH equal to ΔfH?

(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 ΔfH 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, ΔrH equals to heat of formation, ΔfH.

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?

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