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3 years ago

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The reaction 2a + b --> c + d has an activation energy of 80.0 kj/mol. at 320°c, the rate constant k = 1.80 x 10-2 l mol-1 s-

1. what would be the value of k at 450°c?

1 answer:

Naddik [55]3 years ago

5 0

To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:

ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.

ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)

k2 = 0.3325 L / mol-s

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Explanation:

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Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.

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One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume unt

Umnica [9.8K]

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K $0^00C=273K$

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas = 1

$V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L$

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?$

Putting values in above equation, we get:

$\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K$

The final temperature is 900 K

b) The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 6.00 atm

$P_2$ = final pressure of gas = $2\times 6.00atm=12.0atm$

$V_1$ = initial volume of gas = 4.10 L

$V_2$ = final volume of gas = $2\times 4.10 L=8.20L$

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas =?

Now put all the given values in the above equation, we get:

$\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}$

$T_2=1200K$

The final temperature is 1200 K

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