Answer pls ill give brainly

The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the

deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is $0.3115 \ \text{sec}^{-1}$ .

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

$k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}$

$\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}$

where $k$ is the rate constant, $E_{a}$ represents the activation energy of the chemical reaction, $R$ is the gas constant, $T$ is the temperature, and $A$ is the frequency factor.

The frequency factor, $A$ , is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be $k_{1}$ at an initial temperature $T_{1}$ and the rate constant $k_{2}$ at a final temperature $T_{2}$ , thus

$\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)$

$\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}$

Given that $E_{a} \ = \ 26.5 \ \ \text{kJ/mol}$ , $R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}$ , $T_{1} \ = \ \left(40 \ + \ 273\right) \ K$ , $T_{2} \ = \ \left(125 \ + \ 273\right) \ K$ , and $k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}$ , therefore,

$k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}$

8 0

2 years ago

. Which of the following statements about noble gases is NOT true?

WARRIOR [948]

Answer:

d, All of them are found in Earth's atmosphere in small amounts

7 0

2 years ago

Read 2 more answers

What chemical warfare agent (cwa) primarily attacks the airway and lungs, causing irritation of the entire airway from the nose

VLD [36.1K]

The chemical warfareagent that attacks primarily the airway and the lungs would be a choking agent. It would cause irritation of the whole respiratory system resulting to dry-land drowning. Examples are chlorine gas, phosgene, acrolein and chloropicrin.

5 0

3 years ago

An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If

ANTONII [103]

Answer:

The molecular formula of this compound is C4H4N2O2

Explanation:

Step 1: Data given

Mass of the compound = 84 mg

The compound contains:

36 mg of Carbon

3 mg of hydrogen

21 mg of nitrogen

24 mg of oxygen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Molar mass of nitrogen = 14 g/mol

Molar mass of oxygen = 16 g/mol

Step 2: Calculate number of moles

Moles = mass / molar mass

Moles of carbon = 0.036 g/ 12.01 g/mol = 0.003 moles

Moles of hydrogen = 0.003 g/ 1.01 g/mol = 0.003 moles

Moles of nitrogen = 0.021 g/ 14 g/mol = 0.0015 moles

Moles of oxygen = 0.024 g/ 16 g/mol = 0.0015 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.003 / 0.0015 = 2

H: 0.003 / 0.0015 = 2

N = 0.0015/0.0015 = 1

O = 0.0015/0.0015 = 1

The empirical formula is C2H2NO

The molecular mass of this empirical formula is 56 g/mol

Step 4: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 112 g/mol / 56g/mol = 2

We have to multiply the empirical formula by 2

Molecular formula = 2*(C2H2NO) = C4H4N2O2

The molecular formula of this compound is C4H4N2O2

5 0

3 years ago

What is the answer if u know please answer

forsale [732]

An atom having 52 protons and 54 electrons would have an atomic number of 52 and a net charge of -2. This element would be 52 Te 2-, or choice A.

7 0

3 years ago