Skip to content
Question of 98

Q.State and explain Hess law of constant heat summation.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2020Subjective· 4mImportance★★★★★
0% · 0/98 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Hess's Law of Constant Heat Summation: the overall enthalpy change of a reaction is independent of the path/number of steps taken, because enthalpy (H) is a state function.

Statement: If a chemical reaction can be made to occur in several different ways, in one step or through a number of intermediate steps, the total enthalpy change is always the same, irrespective of the path or the number of steps used to bring about the reaction — provided the initial and final conditions (reactants and products) are the same.

Reasoning: Enthalpy (H) is a state function — it depends only on the initial and final states of the system, not on the path taken between them. Since ΔH\Delta H depends only on initial and final states, adding up ΔH\Delta H values for any set of steps that take reactants to the same products must give the same total as the direct single-step ΔH\Delta H.

Illustration:

Consider the formation of CO2 from carbon and oxygen.

Path 1 (direct, one step):

C(s)+O2(g)⟶CO2(g)ΔH=−393.5 kJ mol−1C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} \qquad \Delta H = -393.5\ kJ\,mol^{-1}

Path 2 (via two steps, through CO):

C(s)+12O2(g)⟶CO(g)ΔH1=−110.5 kJ mol−1C_{(s)} + \tfrac{1}{2}O_{2(g)} \longrightarrow CO_{(g)} \qquad \Delta H_1 = -110.5\ kJ\,mol^{-1}

CO(g)+12O2(g)⟶CO2(g)ΔH2=−283.0 kJ mol−1CO_{(g)} + \tfrac{1}{2}O_{2(g)} \longrightarrow CO_{2(g)} \qquad \Delta H_2 = -283.0\ kJ\,mol^{-1}

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.