Consider the following reaction.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) ΔH = -891 kJ
Calculate the enthalpy change for each of the following cases.
(a) 2.00 g methane is burned in excess oxygen.
(b) 2.00 ✕ 103 L methane gas at 733 torr and 22°C is burned in excess oxygen.
I got 111 kJ for part a and 7.10e4 for part b, but it says that both are incorrect.
Could someone help me with this Chemistry Webassign problem?
2017-11-04 11:28 pm
回答 (3)
2017-11-05 12:09 am
✔ 最佳答案
(a)CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) …. ΔH = -891 kJ
The enthalpy change of burning 1 mole of CH₄ (methane) is -891 kJ.
Molar mass of CH₄ = (12.0 + 1.0×4) g/mol = 16.0 g/mol
No. of moles of CH₄ burned = (2.00 g) / (16.0 g/mol) = 0.125 mol
Enthalpy change = (-891 kJ/mol CH₄) × (0.125 mol CH₄) = -111 kJ
(b)
Consider the CH₄ gas :
Pressure, P = 733/760 atm
Volume, V = 2.00 × 10³ L
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 22) K = 295 K
Gas law: PV = nRT
Then, n = PV/(RT)
No. of moles of CH₄ burned, n = (733/760) × (2.00 × 10³) / (0.0821 × 295) mol CH₄
Enthalpy change = (-981 kJ/mol CH₄) × [(733/760) × (2.00 × 10³) / (0.0821 × 295) mol CH₄] = -78100 kJ
2017-11-05 12:07 am
a) Moles CH4 = 2.00 g / 16.0 g/mol = 0.125 mol CH4
-891 kJ/mol X 0.125 mol = -111 kJ
Perhaps you need to include the negative sign since heat is released.
b) I'm guessing that is the problem here, too. To get to the answer, use the ideal gas law to calculate moles methane, multiply that by Delta H, and you are good.
-891 kJ/mol X 0.125 mol = -111 kJ
Perhaps you need to include the negative sign since heat is released.
b) I'm guessing that is the problem here, too. To get to the answer, use the ideal gas law to calculate moles methane, multiply that by Delta H, and you are good.
2017-11-04 11:48 pm
-891 X 2/16 = -111kJ
收錄日期: 2021-05-01 15:12:09
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