At a definite temperature for which of the following will the change (${{\Delta E}} - {{\Delta H}}$) be maximum?
A.${\text{PC}}{{\text{l}}_{\text{5}}} \to {\text{PC}}{{\text{l}}_{\text{3}}}{\text{ + C}}{{\text{l}}_{\text{2}}}$
B.${\text{N}}{{\text{H}}_{\text{4}}}{\text{S}} \to {\text{N}}{{\text{H}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}$
C.${{\text{N}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}} \to 2{\text{NO}}$
D.${\text{CaC}}{{\text{O}}_{\text{3}}} \to {\text{CaO + C}}{{\text{O}}_{\text{2}}}$
Answer
588.9k+ views
Hint: The first law of thermodynamics states that the amount of heat supplied to a body is spent in two ways, firstly to increase the internal energy and secondly to perform external pressure volume work.
Formula used: ${{\Delta H = \Delta E + \Delta }}\left( {{{PV}}} \right){{ = \Delta E + P\Delta V + V\Delta P}}$
where, ${{\Delta H}}$ represents the change in enthalpy in the system and ${{\Delta E}}$ represents the change in the internal energy of the system.
Complete Step by step Solution:
- For the definite temperature we can say that, $\Delta \left( {{\text{PV}}} \right){\text{ = }}\Delta {\text{nRT}}$
where $\Delta {\text{n}}$ represents the change in the number of moles.
- If we have a look at the given equations, then in the first, second, and the fourth equation, the change in the number of moles is positive, because the initial number of the reactants is less than the final no. of moles of the products.
- Hence, in all these cases, the change in enthalpy will be higher than the change in the internal energy. But in equation 3, the number of moles of the reactant is equal to the number of moles of the product, hence $\Delta {\text{n}}$ = 0.
- Therefore, the change (${{\Delta E}} - {{\Delta H}}$) will be maximum as the entire heat supplied will be spent in raising the internal energy of the system.
So, the correct option is C.
Notes: For ideal gases, the combined gas equations that combines the Boyle’s law, Charles’ law, and the Avogadro’s Law, the combined gas law equation can be given as,
${\text{PV = nRT}}$, where the pressure of the gas is ‘P’, the volume of the gas is ‘V’, n is the number of moles of the gas, R is the universal gas constant, and T is the absolute temperature.
Formula used: ${{\Delta H = \Delta E + \Delta }}\left( {{{PV}}} \right){{ = \Delta E + P\Delta V + V\Delta P}}$
where, ${{\Delta H}}$ represents the change in enthalpy in the system and ${{\Delta E}}$ represents the change in the internal energy of the system.
Complete Step by step Solution:
- For the definite temperature we can say that, $\Delta \left( {{\text{PV}}} \right){\text{ = }}\Delta {\text{nRT}}$
where $\Delta {\text{n}}$ represents the change in the number of moles.
- If we have a look at the given equations, then in the first, second, and the fourth equation, the change in the number of moles is positive, because the initial number of the reactants is less than the final no. of moles of the products.
- Hence, in all these cases, the change in enthalpy will be higher than the change in the internal energy. But in equation 3, the number of moles of the reactant is equal to the number of moles of the product, hence $\Delta {\text{n}}$ = 0.
- Therefore, the change (${{\Delta E}} - {{\Delta H}}$) will be maximum as the entire heat supplied will be spent in raising the internal energy of the system.
So, the correct option is C.
Notes: For ideal gases, the combined gas equations that combines the Boyle’s law, Charles’ law, and the Avogadro’s Law, the combined gas law equation can be given as,
${\text{PV = nRT}}$, where the pressure of the gas is ‘P’, the volume of the gas is ‘V’, n is the number of moles of the gas, R is the universal gas constant, and T is the absolute temperature.
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