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प्रश्न
Answer the following question.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
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उत्तर
1. Hess’s law of constant heat summation:
Hess’s law of constant heat summation states that “Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction”.
2. Illustration:
- 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 the reaction, reactants and products in the individual steps are added or subtracted like algebraic entities.
- Consider the synthesis of NH3,
- \[\ce{2H_{2(g)} + N_{2(g)} → N2H_{4(g)}, Δ_r H^0_1 = + 95.4 kJ}\]
- \[\ce{N2H_{4(g)} + H_{2(g)} → 2NH_{3(g)}, Δ_r H^0_2 = -187.6 kJ}\]
\[\ce{H_{2(g)} + N_{2(g)} → 2NH_{3(g)}, Δ_rH^0 = - 92.2 kJ}\]
The sum of the enthalpy changes for steps (1) and (2) is equal to the enthalpy change for the overall reaction.
3. Application of Hess’s law:
The Hess's law has been useful to calculate the enthalpy changes for the reactions with their enthalpies being not known experimentally. To calculate heat of formation, combustion, neutralization, ionization, etc.
संबंधित प्रश्न
Answer the following in one or two sentences.
What is standard state of a substance?
The standard enthalpy of formation of water is - 286 kJ mol-1. Calculate the enthalpy change for the formation of 0.018 kg of water.
State and explain Hess’s law of constant heat summation.
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.
Define the Bond enthalpy.
Define the Enthalpy of ionisation.
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
When 6.0 g of graphite reacts with dihydrogen to give methane gas, 37.4 kJ of heat is liberated. What is standard enthalpy of formation of CH4 (g)?
Daily requirement of energy of a person is 'x' kJ. If heat of combustion of food material (Molecular mass = 100 g) is 'y' kJ, his daily consumption of the food in gram would be ____________.
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?
Which of the following compounds is Not present in its standard state at 25°C and 1 atmosphere pressure?
Given the reaction,
\[\ce{CH2O_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}\] ΔH = −527 kJ
How much heat will be evolved in the formation of 60 g of CO2?
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 ____________.
Given that,
\[\ce{C_{(s)} + O_{2(g)} -> CO_{2(g)}}\] ΔH° = -X kJ
\[\ce{2CO_{(g)} + O_{2(g)} -> 2CO_{2(g)}}\] ΔH° = - Y kJ, then standard enthalpy of formation of carbon monoxide 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?
Standard entropies of N2(g), H2(g), and NH3(g) are a1, a2 and a3 J K-1 mol-1 respectively. What is value of ΔS° for formation of NH3(g)?
Which of the following equations has ΔfH° and ΔH° same?
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
- \[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}ΔH^° = - 726 kJ mol^{-1}}\]
- \[\ce{C_{(s)} + O2_{(g)} → CO2_{(g)}Δ_cH^° = – 393 kJ mol^{-1}}\]
- \[\ce{H2_{(g)} + 1/2 O2_{(g)} -> H2O_{(l)}Δ_fH^° = - 286 kJ mol^{-1}}\]
From the following bond energies:
H – H bond energy: 431.37 kJ mol−1
C = C bond energy: 606.10 kJ mol−1
C – C bond energy: 336.49 kJ mol−1
C – H bond energy: 410.50 kJ mol−1
Enthalpy for the given reaction will be:
\[\begin{array}{cc}
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\ce{C = C + H - H -> H - C - C - H}\\
\phantom{.}|\phantom{....}|\phantom{....................}|\phantom{....}|\phantom{.....}\\
\phantom{}\ce{H}\phantom{...}\ce{H}\phantom{...................}\ce{H}\phantom{...}\ce{H}\phantom{....}
\end{array}\]
Which of the following reactions defines the enthalpy of formation?
The enthalpy of combustion of S (rhombic) is − 297 kJ mo1-1. Calculate the amount of sulphur required to produce 29. 74 kJ of heat.
Draw energy profile diagram and show:
- activated complex
- energy of activation for forward and backward reactions
- enthalpy of reaction
Calculate heat evolved for combustion of 13 gm of acetylene (C2H2).
Given: \[\ce{C2H2_{(g)} + 5/2O_{2(g)}-> 2CO_{2(g)} + H2O_{(l)} \Delta_{(c)}H^{0} = - 1300 kJ}\]
