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3 years ago

The following questions pertain to a system contains 122 g CO(g) in a 0.400 L container at -71.2 degrees C.

Chemistry

1 answer:

7nadin3 [17]3 years ago

8 0

Answer:

a. P = 182 atm

Explanation:

Data Given:

amount of CO = 122g

Volume of CO = .400 L

Temperature of CO = -71.2°C

Convert the temperature to Kelvin

T = °C + 273

T = -71.2 + 273

T = 201.8 K

a. Calculate the pressure exerted by the CO(g) in this system using the ideal gas equation (P) = ?

Solution:

To calculate Pressure by using ideal gas formula

PV = nRT

Rearrange the equation for Pressure

P = nRT / V . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant = 0.08206 L.atm / mol. K

For this we have to know the mole of the gas and the following formula will be used

no. of moles = mass in grams / molar mass . . . . . . (2)

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in equation 2

no. of moles = 122 g / 28 g/mol

no. of moles = 4.4 mol

Now put the value in formula (1) to calculate Pressure for CO

P = 4.4 x 201.8 K x 0.08206 (L.atm/mol. K) / 0.400 L

P = 182 atm

So the pressure will be 182 atm

__________

b. Data Given:

Actual pressure exerted by CO = 145 atm

expected pressure exerted by CO = 182 atm

why the actual pressure is less than what would be expected = ?

Explanation:

This is because of the deviation from ideal behavior of real gases.

The real gases approach to ideal behavior under very high temperature and very low pressure.

But CO deviate from ideal behavior to give expected value for pressure, because it behave at high pressure and low temperature.

This non-ideal behavior is due to two postulate of ideal behavior

gas molecules have negligible volume
Gas molecules have negligible inter-molecular interaction

but these postulates not obeyed under real condition. so we calculated the pressure using ideal condition values for gas and obtained the expected value for pressure but the actual pressure value was detected under normal condition.

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The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

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Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

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$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

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$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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