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# 3- Find the temperature at which Kp 42-0 for the reaction HIR) + 11g).docx

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# 3- Find the temperature at which Kp 42-0 for the reaction HIR) + 11g).docx

3. Find the temperature at which Kp 42.0 for the reaction HIR) + 11g) ?2Hug). [Given: at 25\"C, for HIgÂ·K-0. 1K: 13 1.0 1/ 01-K; for l g), ??\'-62 26 k/ mol, S. = 260.6 / nol ; for HI(g assume that ??. and as. are independent of temperature). (10 points) ??.,-25.9 k/ ol, S. 20631/ 4. Given the data: (10 points) NO(g), ??:-+90.4 kJ mor, s\'-+2 10.6 J mork\' Calculate the standard free energy change, AG, for the reaction: 2 NHylg)+5/2 0+49) 2 NO(g)+3 H2On
Solution
3. For the reaction,
dHo = dHo(products) - dHo(reactants)
= (2 x 25.9) - (0 + 62.26)
= -10.46 kJ/mol = -10460 J/mol
dSo = dSo(products) - dSo(reactants)
= (2 x 206.3) - (131 + 260.6)
= 21 J/K.mol
Kp = 42
dGo = dHo - TdSo = -RTlnKp
-10460 - T(21) = -8.314 x T ln(42)
-10460 - T(21) = -31.075T
T = -10460/-10.075 = 1038.21 K
So, at T = 1038.21 K, Kp = 42
--
4. For the reaction,
dHo = dHo(products) - dHo(reactants)
= (2 x 90.4 + 3 x -286) - (2 x -46 + 5/2 x 0)
= -585.2 kJ/mol
dSo = dSo(products) - dSo(reactants)
= (2 x 210.6 + 3 x 69.96) - (2 x 192.5 + 5/2 x 205)
= -266.42 J/K.mol
T = 273 + 25oC = 298 K
So standard free energy change for the reaction dGo,
dGo = dHo - TdSo
= -585.2 - 298 x -0.26642
= -505.81 kJ/mol
--
5. molar entropy of vaporization dS
dS = dH/Tb
= 21.16 kJ/mol x 1000/(-35.4 + 273)
= 89.06 J/K.mol
--
6.
Ksp for Fe(OH)2 = 1.6 x 10^-14
with x amount of Fe(OH)2 in solution, we would have x amount of Fe2+ and 2x amount of OH-
Ksp = [Fe2+][OH-]^2
1.6 x 10^-14 = (x)(2x)^2
molar solubility = x = 1.60 x 10^-5 M
[OH-] = 2 x 1.60 x 10^-5 M = 3.2 x 10^-5 M
pOh = -log(3.2 x 10^-5) = 4.5
pH = 14 - 4.5 = 9.5
.

3. Find the temperature at which Kp 42.0 for the reaction HIR) + 11g) ?2Hug). [Given: at 25\"C, for HIgÂ·K-0. 1K: 13 1.0 1/ 01-K; for l g), ??\'-62 26 k/ mol, S. = 260.6 / nol ; for HI(g assume that ??. and as. are independent of temperature). (10 points) ??.,-25.9 k/ ol, S. 20631/ 4. Given the data: (10 points) NO(g), ??:-+90.4 kJ mor, s\'-+2 10.6 J mork\' Calculate the standard free energy change, AG, for the reaction: 2 NHylg)+5/2 0+49) 2 NO(g)+3 H2On
Solution
3. For the reaction,
dHo = dHo(products) - dHo(reactants)
= (2 x 25.9) - (0 + 62.26)
= -10.46 kJ/mol = -10460 J/mol
dSo = dSo(products) - dSo(reactants)
= (2 x 206.3) - (131 + 260.6)
= 21 J/K.mol
Kp = 42
dGo = dHo - TdSo = -RTlnKp
-10460 - T(21) = -8.314 x T ln(42)
-10460 - T(21) = -31.075T
T = -10460/-10.075 = 1038.21 K
So, at T = 1038.21 K, Kp = 42
--
4. For the reaction,
dHo = dHo(products) - dHo(reactants)
= (2 x 90.4 + 3 x -286) - (2 x -46 + 5/2 x 0)
= -585.2 kJ/mol
dSo = dSo(products) - dSo(reactants)
= (2 x 210.6 + 3 x 69.96) - (2 x 192.5 + 5/2 x 205)
= -266.42 J/K.mol
T = 273 + 25oC = 298 K
So standard free energy change for the reaction dGo,
dGo = dHo - TdSo
= -585.2 - 298 x -0.26642
= -505.81 kJ/mol
--
5. molar entropy of vaporization dS
dS = dH/Tb
= 21.16 kJ/mol x 1000/(-35.4 + 273)
= 89.06 J/K.mol
--
6.
Ksp for Fe(OH)2 = 1.6 x 10^-14
with x amount of Fe(OH)2 in solution, we would have x amount of Fe2+ and 2x amount of OH-
Ksp = [Fe2+][OH-]^2
1.6 x 10^-14 = (x)(2x)^2
molar solubility = x = 1.60 x 10^-5 M
[OH-] = 2 x 1.60 x 10^-5 M = 3.2 x 10^-5 M
pOh = -log(3.2 x 10^-5) = 4.5
pH = 14 - 4.5 = 9.5
.

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### 3- Find the temperature at which Kp 42-0 for the reaction HIR) + 11g).docx

1. 1. 3. Find the temperature at which Kp 42.0 for the reaction HIR) + 11g) ?2Hug). [Given: at 25"C, for HIgÂ·K-0. 1K: 13 1.0 1/ 01-K; for l g), ??'-62 26 k/ mol, S. = 260.6 / nol ; for HI(g assume that ??. and as. are independent of temperature). (10 points) ??.,-25.9 k/ ol, S. 20631/ 4. Given the data: (10 points) NO(g), ??:-+90.4 kJ mor, s'-+2 10.6 J mork' Calculate the standard free energy change, AG, for the reaction: 2 NHylg)+5/2 0+49) 2 NO(g)+3 H2On Solution 3. For the reaction, dHo = dHo(products) - dHo(reactants) = (2 x 25.9) - (0 + 62.26) = -10.46 kJ/mol = -10460 J/mol dSo = dSo(products) - dSo(reactants) = (2 x 206.3) - (131 + 260.6) = 21 J/K.mol Kp = 42 dGo = dHo - TdSo = -RTlnKp -10460 - T(21) = -8.314 x T ln(42) -10460 - T(21) = -31.075T T = -10460/-10.075 = 1038.21 K So, at T = 1038.21 K, Kp = 42 --
2. 2. 4. For the reaction, dHo = dHo(products) - dHo(reactants) = (2 x 90.4 + 3 x -286) - (2 x -46 + 5/2 x 0) = -585.2 kJ/mol dSo = dSo(products) - dSo(reactants) = (2 x 210.6 + 3 x 69.96) - (2 x 192.5 + 5/2 x 205) = -266.42 J/K.mol T = 273 + 25oC = 298 K So standard free energy change for the reaction dGo, dGo = dHo - TdSo = -585.2 - 298 x -0.26642 = -505.81 kJ/mol -- 5. molar entropy of vaporization dS dS = dH/Tb = 21.16 kJ/mol x 1000/(-35.4 + 273) = 89.06 J/K.mol -- 6. Ksp for Fe(OH)2 = 1.6 x 10^-14 with x amount of Fe(OH)2 in solution, we would have x amount of Fe2+ and 2x amount of OH- Ksp = [Fe2+][OH-]^2 1.6 x 10^-14 = (x)(2x)^2
3. 3. molar solubility = x = 1.60 x 10^-5 M [OH-] = 2 x 1.60 x 10^-5 M = 3.2 x 10^-5 M pOh = -log(3.2 x 10^-5) = 4.5 pH = 14 - 4.5 = 9.5