When a mixture of oxygen and hydrogen is ignited in a test tube, the

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

What mass of KCl is contained in 3.00 L of solution that has a molarity of 1.50M

Soloha48 [4]

Answer:

335.25g

Explanation:

using equation of molarity

8 0

3 years ago

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What is the density (g/mL) of an object that has a mass of 0.03 kg and occupies a volume of 25 mL?

tatyana61 [14]

First convert the kg to g ----- 0.03kg = 30g
Then divide the mass by the volume ----- 30g ÷ 25mL = 1.2
The density is 1.2g/mL<span />

5 0

3 years ago

Read 2 more answers

Test questions Accidently did this

PSYCHO15rus [73]

I don’t see any questions

6 0

3 years ago

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How carbon dioxide prepared in lab?

Nana76 [90]

Explanation:

In laboratory, carbon dioxide is not obtained from the air, carbon dioxide is prepared from carbonates by adding acids to the carbonates. In this experiment carbon dioxide is prepared by the reaction of marble with hydrochloric acid.

7 0

3 years ago