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fenix001 [56]
3 years ago
6

A 6.50x10^-5 m solution of potassium permanganate has a percent transmittance of 27.3% when measured in a 1.15 cm cell at a wave

length of 525 nm. calculate the absorbance of this solution
Chemistry
1 answer:
mihalych1998 [28]3 years ago
7 0
For this problem, we use the Beer Lambert's Law. Its usual equation is:

A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
C is the concentration of the sample solution

As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:

A = 2 - log(%T)
A = 2 - log(27.3)
A = 0.5638
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ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea
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Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

<u>Step 1: </u>Data given

H2(g) + I2(g) ⇌ 2HI(g)     ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

<u>Step 2</u>: Calculate ΔG

ΔG = ΔG° + RTln Q  

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q  

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)  

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

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