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Amanda [17]
2 years ago
11

Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)

nh4cl(s) → nh3(g) + hcl(g)
Chemistry
2 answers:
steposvetlana [31]2 years ago
6 0
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
Mamont248 [21]2 years ago
5 0

Answer:

ChemistryHelper2024's answer is correct, EXCEPT b) is actually -48.3.

Explanation:

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In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
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solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

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Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

                                                       =0.045 gms


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2 years ago
If a dozen apples has a mass of 2.0 kg and 0.20 bushel is 1 dozen apples, How many bushels of apples are in 1.0 Kg of apples
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If 1 dozen apples has a mass of 2.0 kg and 0.20 bushel is 1 dozen apples, how many bushels of apples are in 1.0 kg of apples?

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