Chemistry - Thermochemistry Question with Solution | TestHub
An intimate mixture of hydrogen gas and the theoretical amount of air at 25°C and a total pressure of 1 atm, is exposed in a closed rigid vessel. If the process occurs under adiabatic condition, then using the following data, answer the following questions:
Given: (i) CP,m = 8.3 cal/K-mol (ii) CP,m = 11.3 cal/K-mol (iii) Δf H [H2O(g)] = -57.8 kcal (iv) Air contains 20% O2 and 80% N2 by volume
What will be the maximum temperature (approximately) attained if the process occurs in adiabatic container?
Options:
Answer:
Solution:
Consider the reaction: 2H2 + O2 + 4N2 tarrow 2H2O(g) + 4N2.
(Note: In air, 1 mol of O₂ is accompanied by 4 mol of N₂).
The enthalpy change for this reaction is Δ H = 2(-57.8) = -115.6 kcal.
The number of gas moles changes from 7 to 6, so Δ n = -1.
The change in internal energy, Δ U, can be calculated using the formula Δ U = Δ H - Δ n RT.
Given R = 2 cal/K-mol and assuming T = 298 K:
Δ U = -115.6 + (2 × 298)/1000 = -115.6 + 0.596 = -115.004 kcal.
For a rigid and adiabatic process, this heat raises the temperature of the products at constant volume.
The molar heat capacity at constant volume, CV,m, is related to the molar heat capacity at constant pressure, CP,m, by CV,m = CP,m - R.
Given R = 2 cal/K-mol, the specific heat capacities are:
CV(N2) = 6.3 cal/K-mol
CV(H2O) = 9.3 cal/K-mol
The total heat capacity at constant volume for the products is:
CV,total = 2 × CV(H2O) + 4 × CV(N2)
CV,total = 2(9.3) + 4(6.3) = 18.6 + 25.2 = 43.8 cal/K.
The temperature change, Δ T, can be calculated using Δ U = CV,total Δ T.
Δ T = 115000/43.8 ≈ 2626 K.
Therefore, the final temperature T ≈ 2.9 × 103 K.
(Using Δ H instead of Δ U would yield Δ T ≈ 2937 K).
The closest option is (a).
