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katrin [286]
3 years ago
5

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant.

Chemistry
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

The correct option is: A. 0.168 M

Explanation:

Chemical reaction involved:

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) → 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O

Given: <u>For MnO₄⁻ solution</u>-

Number of moles: n₁ = 1, Volume: V₁ = 20.2 mL, Concentration: M₁ = 0.0250 M;

<u>For Fe²⁺ solution</u>:

Number of moles: n₂ = 5, Volume: V₂ = 15 mL, Concentration: M₂ = ?M

<u><em>To find out the concentration of Fe²⁺ solution (M₂), we use the equation:</em></u>

\frac{M_{1}\times V_{1}}{n_{1}} = \frac{M_{2}\times V_{2}}{n_{2}}

\frac{0.0250 M\times 20.2 mL}{1} = \frac{M_{2}\times 15 mL}{5}

0.505 = M_{2}\times 3

M_{2} = \frac{0.505}{3} = 0.168M

<u>Therefore, the concentration or molarity of Fe²⁺ solution: </u><u>M₂ = 0.168 M</u>

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

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