The reaction for the formation of hydrogen iodide does not go to completion but reaches equilibrium: H2 + I2 = 2HI?
2017-10-02 11:34 pm
A mixture of 1.9 mol of H2 and 1.9 mol of I2 was prepared and allowed to reach equilibrium at 30 atm. The resulting equilibrium mixture was found to contain 3 mol of HI. Calculate the value of Kp.
回答 (1)
2017-10-02 11:50 pm
___________ H₂(g) ____ + ____ I₂(g) ____ ⇌ ____ 2HI(g) …… Kp
Initial: ____ 1.9 mol ________ 1.9 mol ________ 0 mol
Change: ___ -y mol ________ -y mol ________ +2y mol
At eqm: _ (1.9 - y) mol ____ (1.9 - y) mol ______ 2y mol
At eqm: n(HI) = 2y mol = 3 mol, and thus y = 1.5
Hence, n(H₂) = n(I₂) = (1.9 - 1.5) mol = 0.4 mol
Total number of moles in the system = (3 + 0.4*2) = 3.8 mol
At eqm:
P(H₂) = P(I₂) = 300 * (0.4/3.8) atm = 120/3.8 atm
P(HI) = 300 * (3/3.8) atm = 900/3.8 atm
Kp = (900/3.8)² / (120/3.8)² = 900² / 120² = 56
Initial: ____ 1.9 mol ________ 1.9 mol ________ 0 mol
Change: ___ -y mol ________ -y mol ________ +2y mol
At eqm: _ (1.9 - y) mol ____ (1.9 - y) mol ______ 2y mol
At eqm: n(HI) = 2y mol = 3 mol, and thus y = 1.5
Hence, n(H₂) = n(I₂) = (1.9 - 1.5) mol = 0.4 mol
Total number of moles in the system = (3 + 0.4*2) = 3.8 mol
At eqm:
P(H₂) = P(I₂) = 300 * (0.4/3.8) atm = 120/3.8 atm
P(HI) = 300 * (3/3.8) atm = 900/3.8 atm
Kp = (900/3.8)² / (120/3.8)² = 900² / 120² = 56
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