When might a plot of ln[a] vs time not yield a straight line?

1 answer:

6 0

Considering a reaction:
A → B
The rate equation may be described as:
r = -k[A]ⁿ
Taking the natural log,
ln(r) = -nln([A]) + ln(k)
Therefore, the only time the graph of ln[A] vs time will be a straight line is when the order of the reaction is 0, meaning the reaction is independent of reactant concentration.

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A solution is made by mixing 36. g of heptane (C_7H_16) and 16. g of chloroform (CHCl_3). Calculate the mole fraction of heptane

Diano4ka-milaya [45]

Answer:

Mole fraction of solute (heptane) → 0.73

Explanation:

Mole fraction = Moles of solute or solvent / Total moles

Let's calculate the moles of everything:

Moles of solute → Mass of solute / Molar mass

36 g / 100 g/mol = 0.36 moles

Moles of solvent → Mass of solvent / Molar mass

16 g / 119.35 g/mol = 0.134 moles

Total moles = 0.36 + 0.134 = 0.494 moles

Mole fraction of solute = 0.36 / 0.494 → 0.73

8 0

4 years ago

A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture

Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

$n=\frac{PV}{RT}$

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

$n=(\frac{1.45*2.50}{0.0821*293})$

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = $8.60g(\frac{1mol}{44g})$