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Fofino [41]
3 years ago
15

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?
Chemistry
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:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

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moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

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.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

which is  53 mL.

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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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3 years ago
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