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Hess Law of Constant Heat Summation Questions

G.H. Hess, a Russian chemist, established a law about the heat of reactions based on experimental observations in 1840. This law is known as Hess’s law after his name. Hess’s law follows from the fact that enthalpy is a state function.

Definition: The total amount of heat evolved or absorbed in a reaction is the same whether it occurs in one step or several steps.

Hess Law of Constant Heat Summation Chemistry Questions with Solutions

Q-1: Which of the following are Hess’s law applications?

a) Enthalpy of Formation Calculations

b) Enthalpy Change Prediction

c) Enthalpy of Allotropic Transformation Calculations

d) All of the above

Answer: d) All of the above

Q-2: Calculate the enthalpy of formation of carbon monoxide (CO) from the following data:

i) C(s)+O2 (g) → CO2(g) ; ΔH= -393.3 kJ/mol

ii) CO +½ O2 (g) → CO2(g) ; ΔH= -282.8 kJ/mol

Answer: We aim at

C(s)+ ½ O2 (g) → CO (g) ; ΔH=?

Subtracting equation ii) from i), we get

C(s) + ½ O2 (g) → CO (g)

Therefore, ΔH= -393.3 -(-282.8) kJ/mol = -110.5 kJ/mol

Q-3: The thermite reaction used for welding of metals involves the reaction

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

What is ΔHo at 25oC for this reaction? Given that the standard heats of formation of Al2O3 and Fe2O3 are -1675.7 kJ and -828.4 kJ/mol, respectively.

Answer: We are given

i) 2Al (s) + 3/2O2 (g) → Al2O3 (s) ; ΔH= -1675.7 kJ/mol

ii) 2Fe (s) + 3/2O2 (g) → Fe2O3 (s) ; ΔH= -828.4 kJ/mol

We aim at

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s) ; ΔH=?

Equation (i) – Equation (ii) gives

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

ΔH = -1675.7 -(-828.4) kJ/mol = -847.3 kJ/mol

Q-4: CH4 +½O2 → CH3OH has a negative enthalpy. Which of the following relationships is correct if the enthalpy of combustion of CH4 and CH3OH is z and y, respectively?

a) z>y

b) z<y

c) z=y

d) Information insufficient

Answer: b) z<y

Explanation:

i) CH4 + 2O2 → CO2 + 2H2O ; ΔH=z

ii) CH3OH + 3/2O2 → CO2 + 2H2O ; ΔH= y

(i)-(ii) gives the required equation, that is, CH4 + ½O2 → CH3OH

Thus, for the given reaction, ΔH = z-y = -ve

This implies, z<y.

Q-5: Hess’ law states that the total amount of heat evolved or absorbed is independent of _______.

a) The nature of the initial reactants

b) The nature of the final products

c) The path taken

d) All of the above

Answer: c) The path taken.

Q-6: How are thermochemical equations advantageous in Hess law?

Answer: Because thermochemical equations can be treated as algebraic equations that can be added, subtracted, multiplied or divided, they are advantageous in Hess’ law.

Q-7: If for (i) C + O2 → CO2

(ii) C +½O2 → CO

(iii) CO + ½O2 → CO2, the heats of reaction are Q, -12 and -4, respectively. Then Q = _______.

a) -2

b) 2

c) -22

d) -16

Answer: d) -16

Explanation: Adding equations ii) and iii) gives the required equation i) for which we have to calculate Q.

Thus, Q = -12-4 = -16.

Q-8: Calculate the change in enthalpy for the reaction 2H2O2 → 2H2O + O2, if the heat of formation for H2O2 and H2O are -190 kJ/mol and -286 kJ/mol, respectively.

Answer: ΔHo(reaction) = [ 2ΔHof(H2O) + ΔHof(O2)]- [2ΔHof(H2O2)] = [2(-286) + (0)] – [2(-190)] = -192 kJ.

Q-9: The standard enthalpy of combustion at 25oC of hydrogen, cyclohexene and cyclohexane are -241, -3800 and -3920 kJ/mol, respectively. Calculate the heat of hydrogenation for cyclohexene.

Answer:

i) H2 + ½ O2 → H2O ; ΔH= -241kJ

ii) C6H10 + 17/2 O2 → 6CO2 + 5H2O ; ΔH= -3800 kJ

iii) C6H12 + 9O2 → 6CO2 + 6H2O ; ΔH= -3920 kJ

We aim at C6H10 + H2 → C6H12; ΔH=?

Equation (i) + Equation (ii) – Equation (iii) gives the required result.

Thus, ΔH= -241+(-3800)-(-3920) = -121 kJ

Q-10: The heat of formation of

i) CO2(g) from its elements is +94.4 kcal

ii) CuO(s) from its elements is 151.8 kcal and

iii) The heat of reaction between CuO(s) and CO2(g) is +42.25 kcal. Calculate the heat of formation of CuCO3(s).

Answer: We are given

i) C(s) + O2(g) → CO2(g) ; ΔH= 94.4 kcal

ii) Cu(s) + ½ O2 (g) →CuO(s) ; ΔH= 151.8 kcal

iii) CuO(s) + CO2(g) → CuCO3(s) ; ΔH= 42.25 kcal

We aim at

Cu(s) + C(s) + 3/2 O2 → CuCO3(s); ΔH= ?

Eqn(i) + Eqn(ii) + Eqn(iii) gives the required result.

ΔH= 94.4 + 151.8 + 42.25 = 288.45 kcal.

Q-11: What is the name of the law that states that the total enthalpy change in a chemical reaction remains constant at the same temperature?

Answer: Hess’s Law of constant heat summation.

Q-12: What do you mean by a thermochemical equation? Can we use fractional coefficients in such an equation?

Answer: A thermochemical equation is one in which a balanced chemical equation indicates not only the quantities of the different reactants and products but also the amount of heat evolved or absorbed.

Yes, a thermochemical equation can be written using fractional coefficients.

Q-13: ΔHof of CO2(g), CO(g), N2O(g) and NO2(g) in kJ/mol are respectively -393, -110,-81 and -34. Calculate ΔH in kJ for the following reaction.

2NO2(g) + 3CO (g) → N2O (g) + 3CO2 (g)

Answer: ΔHo(reaction) = [ΔHof(N2O) + 3ΔHof(CO2)] -[3(ΔHof(CO) + 2(ΔHof(NO2)]

On substituting values, we get

ΔHo(reaction) == [-81 +3(-393)]-[3(-110)+2(-34)] = -836 kJ

Q-14: What is the basis of Hess’s law?

Answer: Hess’ law is based on the fact that enthalpy is a state function, which means that enthalpy change depends only on the initial and final states and not on the path taken.

Q-15: What is the significance of bond enthalpy?

Answer: One important significance of a bond enthalpy is in calculating the enthalpy of atom formation.

Practice Questions on Hess Law of Constant Heat Summation

Q-1: Hess’s law is concerned with the ________.

a) Entropy of the System

b) Heat Change of the Reaction

c) Kinetic Rate of Reaction

d) Thermodynamics of the Reaction

Q-2: There are two crystalline forms of PbO: one yellow and one red. These two forms have standard enthalpies of formation of -217.3 and -219.0 kJ per mol, respectively. Determine the enthalpy change during the solid-solid phase transition.

PbO (yellow) → PbO (red)

Q-3: Graphite and diamond have combustion enthalpies of 393.5 and 395.4 kJ, respectively. Calculate the enthalpy change associated with the 1 mole of graphite transformed into diamond.

Q-4: Why is the Hess law significant?

Q-5: According to Hess’s law, a chemical reaction is independent of the path taken by chemical reactions while maintaining the same _______.

(a) Initial and final conditions

(b) Only the initial conditions

(c) Only the final conditions

(d) Temperature

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