Unit 8 HW Answers.

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1 Unit 8 HW Answers

2 5.9 a. ΔHA = ΔHB + ΔHC sum of the steps equals overall enthalpy
b. ΔHZ = ΔHX + ΔHY (ΔHY is a negative number) c. The sum of the steps equals overall enthalpy

3 5.10 No, the hydrogen and oxygen are not in their most stable form, not in standard state.

4 5.54 a. q = mcΔT q = (63.88)(4.18)(4.6) q = 1.23kJ 1.23kJ / 0.0485mol
b. endothermic, the dissolving crystals are absorbing heat from the water.

5 5.62 3/2 (2 𝐻 2 + 𝑂 2 ⇄ 2 𝐻 2 𝑂) 3/2 (ΔH = ) -1/2 (3 𝑂 2 ⇄ 2 𝑂 3 ) -1/2 (ΔH = ) 3 𝐻 2 + 𝑂 3 ⇄ 3 𝐻 2 𝑂 kJ

6 5.68 a. 1 2 𝐻 2 𝑔 + 1 2 𝐵𝑟 2 𝑙 → 𝐻𝐵𝑟(𝑔) ΔH°f = -36.23 kJ/mol
b. A𝑔 𝑠 𝑁 2 𝑔 𝑂 2 𝑔 → 𝐴𝑔𝑁 𝑂 3 (𝑠) ΔH°f = kJ/mol c. 2𝐹𝑒 𝑠 𝑂 2 𝑔 → 𝐹𝑒 2 𝑂 3 (𝑠) ΔH°f = kJ/mol d. 2C 𝑠 +2 𝐻 2 𝑔 + 𝑂 2 𝑔 → 𝐶 𝐻 3 𝐶𝑂𝑂𝐻(𝑙) ΔH°f = kJ/mol

7 5.70 2 𝐶 4 𝐻 10 𝑙 +13 𝑂 2 𝑔 →8 𝐶𝑂 2 𝑔 𝐻 2 𝑂 𝑙 Hrxn = 10( ) + 8(-393.5) – 2(-147.6) = / 2mol = kJ/mol 5g / 58g/mol = 0.086mol ( kJ/mol) (0.086mol) = kJ

8 5.72 a. 4HBr 𝑔 + 𝑂 2 𝑔 → 2 𝐻 2 𝑂 𝑙 + 2 𝐵𝑟 2 (𝑙) 2( ) - 4(-36.23) = kJ/mol b. 2NaOH 𝑠 + 𝑆𝑂 3 𝑔 → 𝑁𝑎 2 𝑆𝑂 4 𝑠 + 𝐻 2 𝑂(𝑔) ( ) – (2(-425.6) = kJ/mol c. 𝐶𝐻 4 𝑔 +4 𝐶𝑙 2 𝑔 → 𝐶𝐶𝑙 4 𝑙 + 4𝐻𝐶𝑙(𝑔) 4(-92.30) = kJ/mol d. 𝐹𝑒 2 𝑂 3 𝑠 +6𝐻𝐶𝑙 𝑔 → 2𝐹𝑒 𝐶𝑙 3 𝑠 +3 𝐻 2 𝑂(𝑔) 2(-400) + 3( ) - ( (-92.30)) = kJ/mol

9 5.76 10C 𝑠 +4 𝐻 2 𝑔 → 𝐶 10 𝐻 8 (𝑠) 𝐶 10 𝐻 8 𝑔 +12 𝑂 2 𝑔 →10 𝐶𝑂 2 𝑔 + 4 𝐻 2 𝑂 𝑙 -5154 = 10(-393.5) + 4( ) – X -5154 = – – X 75.68kJ

10 19.3 ΔS will be positive since the reaction is increasing in number of moles of particles

11 19.43 a. C2H6(g) b. CO2(g) Both have more atoms than the other option in each pairing.

12 19.49 a. 𝐶 2 𝐻 4 𝑔 + 𝐻 2 𝑔 → 𝐶 2 𝐻 6 𝑔 ΔS° = -120.48 J/(mol K)
b. 𝑁 2 𝑂 4 𝑔 →2 𝑁𝑂 2 𝑔 ΔS° = J/(mol K) c. 𝐵𝑒 (𝑂𝐻) 2 𝑠 →𝐵𝑒𝑂 𝑠 + 𝐻 2 𝑂(𝑔) ΔS° = J/(mol K) d. 2C 𝐻 3 𝑂𝐻 𝑔 +3 𝑂 2 𝑔 → 2 𝐶𝑂 2 𝑠 +4 𝐻 2 𝑂(𝑔) ΔS° = J/(mol K)

13 19.50 a. 𝑁 2 𝐻 4 𝑔 + 𝐻 2 𝑔 →2 𝑁𝐻 3 𝑔 ΔS° = +15.92 J/(mol K)
b. 𝐾 𝑠 + 𝑂 2 𝑔 → 𝐾𝑂 2 𝑠 ΔS° = J/(mol K) c. 𝑀𝑔 (𝑂𝐻) 2 𝑠 +2𝐻𝐶𝑙(𝑔) →𝑀𝑔 𝐶𝑙 2 𝑠 +2 𝐻 2 𝑂(𝑙) ΔS° = J/(mol K) d. 𝐶𝑂 𝑔 +2 𝐻 2 𝑔 → 𝐶𝐻 3 𝑂𝐻 𝑔 ΔS° = J/(mol K)

14 19.54 a. exothermic b. decreasing randomness (entropy)
c. ΔG = ΔH – TΔS = ( ) = kJ d. Not spontaneous (thermodynamically favorable)

15 19.56 a. 2𝐶𝑟 𝑠 +3 𝑂 2 𝑔 →2 𝐶𝑟 2 𝑂 3 𝑠 ΔH = kJ/mol ΔS = J/(mol K) ΔG = kJ/mol b. 𝐵𝑎 𝐶𝑂 3 𝑠 →𝐵𝑎𝑂 𝑠 + 𝐶𝑂 2 𝑔 ΔH = kJ/mol ΔS = J/(mol K) ΔG = kJ/mol c. 2𝑃 𝑔 +10𝐻𝐹(𝑔) →2𝑃 𝐹 5 𝑔 +5 𝐻 2 (𝑔) ΔH = kJ/mol ΔS = -807 J/(mol K) ΔG = kJ/mol d. 𝐾 𝑠 + 𝑂 2 𝑔 → 𝐾𝑂 2 𝑠 ΔH = kJ/mol ΔS = J/(mol K) ΔG = kJ/mol

16 19.58 a. 𝐻 2 𝑔 + 𝐶𝑙 2 𝑔 →2𝐻𝐶𝑙 𝑔 ΔG = -190.54 kJ/mol
b.𝑀𝑔 𝐶𝑙 2 𝑠 +2 𝐻 2 𝑂 𝑙 →𝑀𝑔𝑂 𝑠 + 2𝐻𝐶𝑙(𝑔) ΔG = kJ/mol c.2 𝑁𝐻 3 𝑔 → 𝑁 2 𝐻 4 𝑔 + 𝐻 2 𝑔 ΔG = kJ/mol d. 2𝑁𝑂𝐶𝑙 𝑔 → 2𝑁𝑂 𝑔 + 𝐶𝑙 2 𝑔 ΔG = kJ/mol

17 19.62 a. ΔG = ΔH – TΔS = -844 – 298(-0.165) = -794.8 kJ
b. ΔG = ΔH – TΔS = 572 – 298(0.179) = 518.7kJ ΔH = TΔS 572 = T(0.179) T = K