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Hess Law Questions

Hess law or Hess law of constant heat summation is a relationship in physical chemistry given by German Hess, a Swiss-born Russian chemist. It states that the amount of heat absorbed or evolved during the entire reaction is independent of the steps taken in the reaction.

Definition: Hess law or Hess law of constant heat summation states that the amount of heat absorbed or evolved during the entire reaction is independent of the steps taken in the reaction.

Hess Law Chemistry Questions with Solutions

Q1. The variations in enthalpy that can not be detected per calorimeter can be seen with

(a) Hess law

(b) Ohm law

(c ) Kreb law

(d) Newton law

Answer: (d) The variations in enthalpy that can not be detected in per calorimeter can be seen with the Hess law.

Q2. Hess’s law states that a chemical reaction is independent of the route of chemical reactions while keeping the same

(a) Initial and Final Conditions

(b) Initial conditions only

(c ) Final conditions only

(d) None of the above

Answer: (a) Hess’s law states that a chemical reaction is independent of the route of chemical reactions while keeping the same initial and final conditions.

Q3. The application of the law of thermodynamics to the enthalpy change was discovered by

(a) Newton

(b) Einstein

(c ) Hess

(d) None of the above

Answer: (c ), The application of the law of thermodynamics to the enthalpy change was discovered by Hess.

Q4. If one or more modes bring about a chemical change in one or more steps, then the amount of heat absorbed or evolved during the entire reaction is the same, whichever way was obeyed. This law is known as

(a) Hess law

(b) Ohm law

(c ) Kreb law

(d) Newton law

Answer: (a), If one or more modes bring about a chemical change in one or more steps, then the amount of heat absorbed or evolved during the entire reaction is the same, whichever way was obeyed. This law is known as Hess law.

Q5. The energy change in a chemical reaction and the surroundings at a constant temperature is known as

(a) Enthalpy

(b) Enthalpy profile

(c ) Enthalpy change

(d) Enthalpy Dynamics

Answer: (c ), The energy change in a chemical reaction and the surroundings at a constant temperature is known as enthalpy change.

Q6. Hess’s law deals with

(a) Heat change in a chemical reaction

(b) Influence of pressure on the volume of gas

(c ) Equilibrium constant

(d) Rate of reaction

Answer: (a) Hess’s law deals with the heat change in a chemical reaction.

Q7. What is Hess law?

Answer: Hess law or Hess law of constant heat summation states that the amount of heat absorbed or evolved during the entire reaction is independent of the steps taken in the reaction.

hess law

Q8. Explain the feasibility of Hess law with an example.

Answer: Consider the formation of carbon dioxide. It can occur in two ways.

Method 1: Carbon reacts with oxygen to form carbon dioxide in a single step. It releases 94.3 kcals of heat.

C + O2 → CO2 (ΔH = 94.3 kcal)

Method 2: Carbon reacts with oxygen to form carbon monoxide in the first step. It releases 26.0 kcals of heat.

2 C + O2 → 2 CO (ΔH1 = 26.0 kcal).

Carbon monoxide again reacts with oxygen to form carbon dioxide in the second step. It releases 68.3 kcals of heat.

2 CO + O2 → 2 CO2 (ΔH2 = 68.3 kcal)

Adding the enthalpy of method two, ΔHT = ΔH1 + ΔH2 = 26.0 + 68.3 = 94.3 kcal.

Thus, the net reaction enthalpy of both reactions is the same as that of single-step formation. So, the enthalpy of the reaction does not change on the path followed by the reactants.

Thus, Hess’s law is verified.

Q9. What is enthalpy?

Answer: Enthalpy is a thermodynamic parameter that measures the total heat present in a thermodynamic system where the pressure is constant.

Q10. What is entropy?

Answer: Entropy is the measure of the degree of randomness. It measures the system’s thermal energy per unit temperature, which is elusive for doing useful work.

Q11. Calculate the standard enthalpy of formation of CH3​OH from the following data.

CH3​OH (l) + 2 ​O2 (g) → CO2 (g) + 2 H2​O (l) (ΔH = − 726 kJmpl−1)

C (graphite)​ + O2 (g) → CO2 (g) (ΔH = − 393 kJmol−1)

H2 ​(g) + 2 O2​ (g) → H2​O (l) (ΔH = − 286 kJmol−1)

Answer: Reversing the equation 1.

CO2 (g) + 2 H2​O (l) → CH3​OH (l) + 2 ​O2 (g) (ΔH = + 726 kJmpl−1)

On reversing the reaction, the sign of ΔH also changes.

The second equation is

C (graphite)​ + O2 (g) → CO2 (g) (ΔH = − 393 kJmol−1).

Multiply equation three by 2,

[H2 ​(g) + 2 O2​ (g) → H2​O (l) (ΔH = − 286 kJmol−1) ] X 2

2 H2 ​(g) + 4 O2​ (g) → 2 H2​O (l) (ΔH = − 286 X 2= – 572 kJmol−1)

Add these three equations.

C (graphite) ​+ 2 H2 (g) + ½ O2​ ​(g) → CH3​OH (l)

Enthalpy of formation of CH3​OH = ΔHT

Enthalpy of formation of CH3​OH = ΔH1 + ΔH2 + ΔH3

Enthalpy of formation of CH3​OH = 726 – 393 – 572 kJmol−1

Enthalpy of formation of CH3​OH = – 239 kJmol−1

Q12. The enthalpies of formation CO (g), CO2 ​(g), N2​O (g) and N2​O4​ (g) are − 110, − 393, 81 and 9.7 KJ/mol, respectively. Find the value of ΔH for the reaction.

N2​O4​ (g) + 3 CO → N2​O (g) + 3 CO2 ​(g).

Answer: ΔH ​= 3 ΔCO2 ​+ 3 ΔN2​O ​− 3 ΔCO ​− Δ N2​O4​

​ΔH ​= 3 × (−393) + (81) − 3 × (−110) − (9.7)

ΔH ​= − (777.7) KJ​

Q13. Calculate the enthalpy change for the following reaction:

CH4 ​(g) + 2 O2 ​(g) ⟶ CO2 ​(g) + 2 H2​O (l).

Given that enthalpies of formation of CH4, CO2 and H2​O are 74.8 kJmol−1,− 393.5 kJ mol−1, and − 286 kJmol−1, respectively.

Answer: ΔHo = ΔHo(products) ​− ΔHo(reactants)

ΔHo = [ΔHo(CO2) ​+ 2 X ΔHo(H2O)] − [ΔHo(CH4) + 2 X ΔHo(O2)​]

ΔHo = [− 393.5 + 2 X (−286.2)] − [−74.8 + 2 X 0]

ΔHo = − 393.5 – 572.4 + 74.8

ΔHo = − 891.1 kJ

Q14. Differentiate between enthalpy and entropy.

Answer:

S. No. Enthalpy Entropy
1. Enthalpy is a thermodynamic parameter that measures the total heat present in a thermodynamic system where the pressure is constant. Entropy is a thermodynamic parameter that measures the system’s thermal energy per unit temperature, which is elusive for doing useful work.
2. It is the sum of internal energy and flows energy. It is the measurement of the randomness of molecules
3. It is denoted by the symbol H. It is denoted by the symbol S.
4. Its unit is Joule mol-1. Its unit is Joule K-1.

Q15. Differentiate between intensive and extensive functions.

Answer:

S. No. Intensive Function Extensive Function
1. It does not depend on the mass. It depends on the mass.
2. It cannot be computed. It can be computed.
3. It is used to determine the identity of a system. It can not be used to determine the identity of a system.
4. It can be easily identified. It cannot be easily identified.
5. Example: Melting point, Boiling point, Freezing point, Colour, Lustre, Pressure, Density, Ductility, Conductivity, Odour, etc. Example: Length, Weight, Mass, and Volume

Practise Questions on Hess law

Q1. One mole of a non-ideal gas undergoes a change state (2atm, 3L, 95K) to (4atm, 5L, 245K) with a change of internal energy, ΔU = 30 L atm. What is the change in enthalpy (ΔH) of the process in L atm?

Q2. The combustion of one mole of benzene occurs at 298 K, and 1 atm after combustion, CO2 (g)​ and H2​O J​ are produced, and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formulation ΔHf of benzene. The standard enthalpy of formation of CO2 (g)​ and H2​O is – 393.5 KJ / Mole and – 285.83 KJ / Mole.

Q3. The reaction of cyanamide NH2​CN(s) with dioxygen was carried out in a bomb calorimeter, and △U was found to be −742.7 kJ mol-1 at 298 K.

NH2​CN (g) + 3 / 2 ​O2 ​(g) → N2 ​(g) + CO2 ​(g) + H2​O (l)

Calculate the enthalpy change for the reaction at 298 K?

Q4. The enthalpy of vaporisation of water at 100oC is 40.63 KJ mol−1. What is the value ΔE for this process?

Q5. The bond dissociation enthalpy of H2​, Cl2​ and HCl are 434, 242 and 431 kJ mol−1. What is the enthalpy of the formation of HCl?

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