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3 years ago

What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,

Chemistry

2 answers:

boyakko [2]3 years ago

8 0

The rate constant of a reaction : 8.3 x 10⁻⁴

<h3>Further explanation</h3>

Given

rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1

Required

the rate constant

Solution

For aA + bB ⇒ C + D

Reaction rate can be formulated:

$\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}$

the rate constant : k =

$\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}$

Firlakuza [10]3 years ago

8 0

Answer:

8.3x10^10-4

Explanation:

a p e x

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In a solution with a pH of 3.0, the color of

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Answer:

A) litmus is red

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To answer this question, it can be helpful to have the color charts. Litmus, phenolphthalein and methyl orange are ways to test the pH of a substance.

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3 years ago

What is the acceleration of a jogger who moves in a straight line and increases her speed by 1 mile per hour every 0.5 hours?

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2 years ago

Alice and Bob are experimenting with two moles of neon, a monatomic gas, that starts out at conditions of standard temperature a

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Answer:

29273.178 joules have been added to the gas for the entire process.

Explanation:

The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

Isochoric (Constant volume)

$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) Alice heats the gas at constant volume until its pressure is doubled:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

$T_{2} = 2\times 273.15\,K$

$T_{2} = 546.3\,K$

(ii) Bob further heats the gas at constant pressure until its volume is doubled:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

$c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{v} = 12.471\,\frac{J}{mol\cdot K}$

$c_{p} = \frac{5}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} \right)$

$c_{p} = 20.785\,\frac{J}{mol\cdot K}$

$Q = (2\,mol)\cdot \left[\left(12.471\,\frac{J}{mol\cdot K} \right)\cdot (536.3\,K-273.15\,K)+\left(20.785\,\frac{J}{mol\cdot K} \right)\cdot (1092.6\,K-546.3\,K )\right]$

$Q = 29273.178\,J$

29273.178 joules have been added to the gas for the entire process.

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3 years ago