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. Hessβs 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) Hessβs 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. Hessβs 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) Hessβs 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.
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 β CO2 (ΞH = 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 β 2 CO (ΞH1 = 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 β 2 CO2 (ΞH2 = 68.3 kcal)
Adding the enthalpy of method two, ΞHT = ΞH1 + ΞH2 = 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 CH3βOH from the following data.
CH3βOH (l) + 2 βO2 (g) β CO2 (g) + 2 H2βO (l) (ΞH = β 726 kJmplβ1)
C (graphite)β + O2 (g) β CO2 (g) (ΞH = β 393 kJmolβ1)
H2 β(g) + 2 O2β (g) β H2βO (l) (ΞH = β 286 kJmolβ1)
Answer: Reversing the equation 1.
CO2 (g) + 2 H2βO (l) β CH3βOH (l) + 2 βO2 (g) (ΞH = + 726 kJmplβ1)
On reversing the reaction, the sign of ΞH also changes.
The second equation is
C (graphite)β + O2 (g) β CO2 (g) (ΞH = β 393 kJmolβ1).
Multiply equation three by 2,
[H2 β(g) + 2 O2β (g) β H2βO (l) (ΞH = β 286 kJmolβ1) ] X 22 H2 β(g) + 4 O2β (g) β 2 H2βO (l) (ΞH = β 286 X 2= – 572 kJmolβ1)
Add these three equations.
C (graphite) β+ 2 H2 (g) + Β½ O2β β(g) β CH3βOH (l)
Enthalpy of formation of CH3βOH = ΞHT
Enthalpy of formation of CH3βOH = ΞH1 + ΞH2 + ΞH3
Enthalpy of formation of CH3βOH = 726 – 393 – 572 kJmolβ1
Enthalpy of formation of CH3βOH = – 239 kJmolβ1
Q12. The enthalpies of formation CO (g), CO2 β(g), N2βO (g) and N2βO4β (g) are β 110, β 393, 81 and 9.7 KJ/mol, respectively. Find the value of ΞH for the reaction.
N2βO4β (g) + 3 CO β N2βO (g) + 3 CO2 β(g).
Answer: ΞH β= 3 ΞCO2 β+ 3 ΞN2βO ββ 3 ΞCO ββ Ξ N2βO4β
βΞH β= 3 Γ (β393) + (81) β 3 Γ (β110) β (9.7)
ΞH β= β (777.7) KJβ
Q13. Calculate the enthalpy change for the following reaction:
CH4 β(g) + 2 O2 β(g) βΆ CO2 β(g) + 2 H2βO (l).
Given that enthalpies of formation of CH4, CO2 and H2βO are 74.8 kJmolβ1,β 393.5 kJ molβ1, and β 286 kJmolβ1, respectively.
Answer: ΞHo = ΞHo(products) ββ ΞHo(reactants)β
ΞHo = [ΞHo(CO2) β+ 2 X ΞHo(H2O)] β [ΞHo(CH4) + 2 X ΞHo(O2)β]
ΞHo = [β 393.5 + 2 X (β286.2)] β [β74.8 + 2 X 0]
ΞHo = β 393.5 – 572.4 + 74.8
ΞHo = β 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, ΞU = 30 L atm. What is the change in enthalpy (ΞH) 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)β and H2βO Jβ are produced, and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formulation ΞHf of benzene. The standard enthalpy of formation of CO2 (g)β and H2βO is – 393.5 KJ / Mole and – 285.83 KJ / Mole.
Q3. The reaction of cyanamide NH2βCN(s) with dioxygen was carried out in a bomb calorimeter, and β³U was found to be β742.7 kJ mol-1 at 298 K.
NH2βCN (g) + 3 / 2 βO2 β(g) β N2 β(g) + CO2 β(g) + H2βO (l)
Calculate the enthalpy change for the reaction at 298 K?
Q4. The enthalpy of vaporisation of water at 100oC is 40.63 KJ molβ1. What is the value ΞE for this process?
Q5. The bond dissociation enthalpy of H2β, Cl2β and HCl are 434, 242 and 431 kJ molβ1. What is the enthalpy of the formation of HCl?
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