Advertisements
Advertisements
Question
State and explain Hess’s law of constant heat summation.
Advertisements
Solution
Hess’s law of constant heat summation states that, “Overall the enthalpy change for a reaction is equal to sum of enthalpy changes of individual steps in the reaction”.
Explanation:
- The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy is a state function. The enthalpy change of a reaction depends only on the initial and final states and not on the path by which the reaction occurs.
- To determine the overall equation of reaction, reactants and products in the individual steps are added or subtracted like algebraic entities.
- Consider the synthesis of NH3,
| 1. | \[\ce{2H_{2(g)} + N_{2(g)} -> N2H_{4(g)}}\], | `Delta_"r""H"_1^0` = +95.4 kJ |
| 2. | \[\ce{N2H_{4(g)} + H_{2(g)} -> 2NH_{3(g)}}\], | `Delta_"r""H"_2^0` = −187.6 kJ |
| \[\ce{3H_{2(g)} + N_{2(g)} -> 2NH_{3(g)}}\], | ∆rH° = – 92.2 kJ |
The sum of the enthalpy changes for steps (1) and (2) is equal to enthalpy change for the overall reaction.
APPEARS IN
RELATED QUESTIONS
Answer the following in one or two sentences.
What is standard state of a substance?
Answer in brief.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer the following question.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Calculate the standard enthalpy of combustion of CH4(g) if ΔfH°(CH4) = – 74.8 kJ mol–1, ΔfH°(CO2) = – 393.5 kJ mol–1 and ΔfH°(H2O) = – 285.8 kJ mol–1.
Calculate the standard enthalpy of formation of liquid methanol from the following data:
- \[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\] ∆H° = – 726 kJ mol–1
- \[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\] ∆cH° = – 393 kJ mol–1
- \[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\] ∆fH° = – 286 kJ mol–1
Write an application of Hess’s law.
Does the following reaction represent a thermochemical equation?
\[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(g)}}\], ∆fH° = –900 kJ mol–1
The heat of formations of CO(g) and CO2(g) are −26.4 kcal and −94.0 kcal respectively. The heat of combustion of carbon monoxide will be ____________.
The standard heats of formation in kcal mol−1 of NO2(g) and N2O4(g) are 8.0 and 2.0 respectively. The heat of dimerization of NO2 in kcal is ____________.
\[\ce{2NO2_{(g)} ⇌ N2O4_{(g)}}\]
The enthalpy change accompanying a reaction in which 1 mole of the substance in the standard state reacts completely with oxygen or is completely burnt is called as ____________.
lf, \[\ce{C_{(s)} + O2_{(g)} -> CO2_{(g)}}\], ∆H = x .........(i)
\[\ce{CO_{(g)} + 1/2O2_{(g)} -> CO2_{(g)}}\], ∆H = y .......(ii)
Then, the heat of formation of CO is:
The heat evolved in the combustion of benzene is given by
\[\ce{C6H6 + 7 1/2O2 -> 6CO2_{(g)} + 3H2O_{(l)}}\]; ΔH = −3264.6 kJ
Which of the following quantities of heat energy will be evolved when 39 g C6H6 are burnt?
The enthalpy of formation of nitrogen dioxide is +33.2 kJ mol−1. The enthalpy of the reaction \[\ce{2N2_{(g)} + 4O2_{(g)} -> 4NO2_{(g)}}\]; is ____________.
Standard enthalpy of formation of water is - 286 kJ mol-1. When 1800 mg of water is formed from its constituent elements in their standard states the amount of energy liberated is ______.
Heat of formation of water is - 272 kJ mol-1. What quantity of water is converted to H2 and O2 by 750 kJ of heat?
Which of the following equations has ΔfH° and ΔH° same?
When the enthalpy of combustion of carbon to carbon dioxide is - 360 kJ mol-1, then the enthalpy change for the formation of 18 g of CO2 from carbon and dioxygen at the same temperature in kJ will be ______.
What is enthalpy of formation of NH3 if bond enthalpies as (N ≡ N) = - 941 kJ/mol.
\[\ce{(H - H)}\] = 436 kJ/mol and \[\ce{(N - H)}\] = 389 kJ/mol?
Define and explain the term, enthalpy of reaction.
Which of the following reactions defines the enthalpy of formation?
Draw energy profile diagram and show:
- activated complex
- energy of activation for forward and backward reactions
- enthalpy of reaction
The heat evolved in the combustion of 6.022 x 1021 carbon particles is 3.94 kJ. The heat of combustion of carbon is ______.
Calculate ΔsubH of the H2O from the given data:
\[\ce{H2O_{(s)}->H2O_{(l)},}\] ΔfusH = 6.01kJ mol−1
\[\ce{H2O_{(l)}-> H2O_{(g)},}\] ΔVapH = 45.07 kJ mol−1.
Calculate the standard enthalpy of combustion of methane if the standard enthalpy of formation of methane, carbon dioxide and water are −74.8, −393.5 and −285.8 kJmol−1 respectively.
