Consider the following reaction. CH4(g) + 2 O2(g) - CO2(g) + 2 H20(1) AH = -891...

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Question

Answer 1

Answer #1

Answer: The reaction is given as follows:

CH₄(g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l) ΔH = - 891 kJ

From the above reaction we can say that for combustion of 1 mole of CH_4, ΔH = - 891 kJ

(a) Now, 2.00 g methane = 2/16 mol of CH₄ = 0.125 mol of CH₄ [∵ molar mass of CH₄ = 16]

∴ ΔH for 0.125 mol of CH₄ = - (0.125 × 891) kJ = -111.375 kJ

(b) We know that for ideal gas PV = nRT

Here, we assume CH₄ behave as ideal gas.

Given, P = 743 torr = 743/760 atm = 0.977 atm, V = 2×10³ Lit, R = 0.082 Lit.atm,

T= 25ºC = (273+25) K = 298K, n = ?

n = PV/RT = (0.977×2000) / (0.082×298) = 79.96 ≈ 80 mol

∴ ΔH for 80 mol of CH₄ = - (80 × 891) kJ = -71,280 kJ

(a) The enthalpy change for 2.00 g CH₄ combustion	- 111.375 kJ
(b) The enthalpy change for 2.00×10³ Lit CH₄ combustion at 743 torr and 25ºC	- 71,280 kJ

Answer 2

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