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3 years ago

The rate constant k for a certain reaction is measured at two different temperatures:

Chemistry

1 answer:

bogdanovich [222]3 years ago

8 0

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are $K_1$ and $K_2$ at temperature $T_1$ and $T_2$

By using Arrhenius equation at two different two different temperature,

$Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K ;K_1=4.8*10^8;T_2=273+280=553K ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;$

By putting value of R=2 cal/mole.K

$E_a=5510.265cal/mole;$

By rounding off upto 2 significant figure;

$E_a=5.5Kcal/mole;$

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A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti

Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)$

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

$n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2$

In such a way, the yielded moles of hydrobromic acid and chlorine are:

$n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2$

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

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What is the energy of a moving mass?

satela [25.4K]

Answer:

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2 years ago

You have a gas stored in a rigid container (fixed volume). Explain the relationship between a decrease in temperature, collision

alexandr1967 [171]

Answer:

A decrease in temperature would decrease kinetic energy, therefore decreasing collisions possible.

Explanation:

A gas at a fixed volume is going to have collisions automatically. If you decrease the temperature (same thing as decreasing kinetic energy) you are cooling down the molecules in the container which gives them less energy and "relaxes" them. This decrease in energy causes them to move around much slower and causing less collisions, at a much slower rate. In a perfect world, these collisions do not slow down the molecule but we know that they do, just a very very small unmeasurable amount.

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3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

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3 years ago