Ncert Solutions Chemistry Class 11th
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41. Amount of heat required to vapourize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vapourization or molar enthalpy of vapourization, ΔvapH?.
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40. Define the following:
(i) First law of thermodynamics.
(ii) Standard molar enthalpy of formation.
40. Define the following:
(i) First law of thermodynamics.
(ii) Standard molar enthalpy of formation.
40. (i) First law of thermodynamics: It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant.
?u = q + w
(ii) The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of agg
Upvote 39. In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the
quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive
Upvote 38. Predict the sign of ?S for the following reaction heat
CaCO3 (s) →CaO(s) + CO2(g)
38. Predict the sign of ?S for the following reaction heat
CaCO3 (s) →CaO(s) + CO2(g)
38. Since, there is an increase in the number of moles on the product side so, entropy increases and ?S is positive.
Upvote 37. Hess law states that “If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.”
Or it can be stated as “The change of enthalpy of a rea
Upvote If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
This is a Short Answer Type Questions as classified in NCERT Exemplar
Molar enthalpy change for graphite (ΔH)
= enthalpy change for 1 g x molar mass of C = -20.7×12 = -2.48 x 102 kJ mol-1
Since the sign of ΔH = -ve, it is an exothermic reaction.
Upvote The difference between Cp and Cv can be derived using the empirical relation H = U + pV. Calculate the difference between Cp and Cv for 10 moles of an ideal gas.
The difference between Cp and Cv can be derived using the empirical relation H = U + pV. Calculate the difference between Cp and Cv for 10 moles of an ideal gas.
This is a Short Answer Type Questions as classified in NCERT Exemplar
Given that, Cv = heat capacity at constant volume,
Cp = heat capacity at constant pressure
Difference between Cp and Cv is equal to gas constant (R).
.’. Cp – Cv = nR
Upvote Heat capacity (CP) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?
Heat capacity (CP) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?
This is a Short Answer Type Questions as classified in NCERT Exemplar
For water, molar heat capacity = 18 x Specific heat or Cp = 18 x c
But, specific heat,
C = 4.18 J g-1 K-1 Heat capacity,
Cp = 18 x 4.18 JK-1 = 75.24 JK-1
Upvote 36. Out of diamond and graphite, which has greater entropy? At what temperature entropy of a substance is zero?
36. Out of diamond and graphite, which has greater entropy? At what temperature entropy of a substance is zero?
36. Graphite has greater entropy since it is loosely packed. At absolute zero the entropy of a substance is zero.
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