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Stoichiometry - Engineering

Q1:

One mole of methane undergoes complete combustion in a stoichiometric amount of air. The reaction proceeds as CH4 + 2O2 → CO2 + 2H2O. Both the reactants and products are in gas phase. ΔH°298 = - 730 kJ/mole of methane. If the average specific heat of all the gases/vapour is 40 J/mole.K, the maximum temperature rise of the exhaust gases in °C would be approximately equal to

A 1225

B 1335

C 1525

D 1735

ANS:D - 1735

To find the maximum temperature rise of the exhaust gases, we can use the principle of conservation of energy. The heat released during the combustion reaction is equal to the heat absorbed by the gases, which leads to an increase in temperature. Given:

  • ΔH°298 for the combustion of methane = -730 kJ/mol of methane
  • Average specific heat of all gases/vapor = 40 J/mol·K
First, we need to find the amount of heat released during the combustion of 1 mole of methane. From the balanced chemical equation: CH4+2O2→CO2+2H2OCH4​+2O2​→CO2​+2H2​O The enthalpy change for the combustion of 1 mole of methane is -730 kJ. Now, we need to calculate the amount of heat absorbed by the gases. This can be calculated using the formula: 𝑄=𝑛×𝐶𝑣×Δ𝑇Q=n×Cv​×ΔT where:
  • 𝑄Q is the heat absorbed (in joules)
  • 𝑛n is the number of moles of gas
  • 𝐶𝑣Cv​ is the molar specific heat capacity of the gases (in J/mol·K)
  • Δ𝑇ΔT is the change in temperature (in K)
Since the combustion of 1 mole of methane releases 730 kJ of heat, the heat absorbed by the gases is also 730 kJ. We need to convert this to joules: 730 kJ=730×103 J730kJ=730×103J Now, we'll use the formula to find the change in temperature (Δ𝑇ΔT): 730×103=(1+2+2)×40×Δ𝑇730×103=(1+2+2)×40×ΔT Δ𝑇=730×1035×40ΔT=5×40730×103​ Δ𝑇=730×103200ΔT=200730×103​ Δ𝑇=3650 KΔT=3650K Therefore, the maximum temperature rise of the exhaust gases is approximately 3650 K3650K.