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Question: For \(CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)\)at \(977{}^\circ C\), \(\Delta H=174kJ/mol\); then \(...

For CaCO3(s)→CaO(s)+CO2(g)CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)at 977∘C977{}^\circ C, ΔH=174kJ/mol\Delta H=174kJ/mol; then ΔE\Delta Eis:
A.160 kJ
B. 163.6 kJ
C. 186.4 kJ
D. 180 kJ

Explanation

Solution

relation between enthalpy change and internal energy can be written in the number of moles as, ΔH=ΔE+ΔnRT\Delta H=\Delta E+\Delta nRT, where n are the number of moles. This number of moles is moles of reactants subtracted from moles of products. Number of moles is taken only for gaseous species.

Complete answer:
Enthalpy of any reaction is the heat contained in that system at constant pressure. Enthalpy is the sum of internal energy and pressure, volume. As, H = E + PV, where, H is enthalpy, E is internal energy, P is pressure, and V is volume.
This equation can be written in terms of number of moles. This is done by replacing PV with nRT, from the relation of ideal gas equation, as PV = nRT. So, the equation can be,
H = E + nRT, for a reaction taking place, at constant pressure and temperature, there is change in these values, so the equation is,
ΔH=ΔE+ΔnRT\Delta H=\Delta E+\Delta nRT, we have been given the value of ΔH=174kJ/mol\Delta H=174kJ/mol and temperature of 977∘C977{}^\circ C.
To find number of moles, subtract number of moles of reactants from number of moles of product from the reaction, CaCO3(s)→CaO(s)+CO2(g)CaC{{O}_{3}}(s)\to CaO(s)+C{{O}_{2}}(g)
As we can see, only 1 product is gaseous, while others are solids, so Δn\Delta n = 1 – 0 = 1, putting all the given values in the formula, ΔH=ΔE+ΔnRT\Delta H=\Delta E+\Delta nRT, rearranging for internal energy, we have,
ΔE=ΔH−ΔnRT\Delta E=\Delta H-\Delta nRT
ΔE=(−174×103)−1(8.314)×1250 J\Delta E=(-174\times {{10}^{3}})-1(8.314)\times 1250\,J(converting kJ into J, and Celsius into Kelvin)
ΔE=163.6 kJ\Delta E=163.6\,kJ
Hence, for the given reaction, ΔE\Delta E is 163.6 kJ.

So, option B is correct.

Note:
The conversion of Celsius to Kelvin is 1∘C=273K1{}^\circ C=273K , so 273 is added in the Celsius value. The conversion for 1 kilojoule to joule is 1 kJ = 1000 J. So, the value of kJ for enthalpy is converted to Joules. The final answer is expressed in kilojoules by dividing the value obtained in joules by 1000.