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leonid [27]
3 years ago
14

The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if

the initial concentration of a was 4.25 m, what was the concentration of a (in m) after 180.0 min?
Chemistry
2 answers:
olga nikolaevna [1]3 years ago
7 0
According to the second order formula:
1/[At] = K t + 1/[Ao]
and when we have the K constant =0.0265 & we have t = 180 min & we have the initial concentration of A = 4.25 so by substitution:

1/[At]  = 0.0265 X 180min + 1/4.25 
1/[At] = 5
∴[At] = 1/5 = 0.2 m
andre [41]3 years ago
7 0

Answer : The final concentration was 0.199 M

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = 0.0265M^{-1}min^{-1}

t = time = 180.0 min

[A_t] = final concentration = ?

[A_o] = initial concentration = 4.25 M

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

0.0265\times 180.0=\frac{1}{[A_t]}-\frac{1}{4.25}

[A_t]=0.199M

Therefore, the final concentration was 0.199 M

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Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.
zysi [14]

P = 11.133 atm (purple)

T = -236.733 °C(yellow)

n = 0.174 mol(red)

<h3>Further explanation  </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:  

  • Boyle's law at constant T, P = 1 / V  
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  • Avogadro's law, at constant P and T, V = n  

So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm  

V = volume, liter  

n = number of moles  

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

T = temperature, Kelvin  

To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

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T= 10 +273.15 = 373.15 K

V=5.5 L

n=2 mol

\tt P=\dfrac{nRT}{V}\\\\P=\dfrac{2\times 0.08205\times 373.15}{5.5}\\\\P=11.133~atm

  • Yellow box

V=8.3 L

P=1.8 atm

n=5 mol

\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC

  • Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol

3 0
3 years ago
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