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3 years ago

The equilibrium concentrations were found to be [br2] = 2.2 × 10−3 m, [cl2] = 3.0 × 10−2 m, and [brcl] = 1.5 × 10−2 m. write the

equilibrium expression, and calculate the equilibrium constant for this reaction at this temperature.

Chemistry

1 answer:

kow [346]3 years ago

7 0

$Br_{2} + Cl_{2}---\ \textgreater \ 2BrCl

K_{eq} = \frac{[BrCl]^{2}}{[Br_{2}][Cl_{2}]]} = \frac{[1.5*10^{-2}]^{2}}{[2.2*10^{-3}][3.0*10^{-2}]} =0.34*10=3.4

K_{eq}=3.4$

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A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol

Gekata [30.6K]

Answer:

0.314 s⁻¹

Step-by-step explanation:

Use the Arrhenius equation.

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

Data:

k₁ = 0.0275 s⁻¹; k₂ = ?

Eₐ = 75.5 kJ·mol⁻¹

T₁ = 20 °C = 293.15 K; T₂ = 45°C = 318.15 K

Calculation:

$\ln(\frac{k_{2} }{0.0275}) = (\frac{75 500 }{8.314})(\frac{ 1}{293.15} - \frac{1 }{318.15 })$

$\ln(\frac{k_{2} }{0.0275}) = 9081\times 2.681 \times10^{-4}$

$\ln(\frac{k_{2} }{0.0275}) = 2.434$

$\frac{k_{2} }{0.0275} = \text{e}^{2.434}$

$\frac{k_{2} }{0.0275} = 11.41$

k₂ = 0.0275 × 11.41

k₂ = 0.314 s⁻¹

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3 years ago

Chemical weathering change the ___ of rock

a_sh-v [17]

Answer:

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Explanation:

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Answer:

it would be an element because its an element

Explanation:

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