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Fudgin [204]
3 years ago
12

Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ

ine whether the reaction is first order or second order. determine whether the reaction is first order or second order. this is the first order reaction. this is the second order reaction. submitprevious answers correct part b what is the rate constant?
Chemistry
1 answer:
Whitepunk [10]3 years ago
7 0
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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Answer is: -963,8 kJ.
Q₁ = m(Fe) · C · ΔT₁.
C - specific heat capacity of liquid iron, C(Fe) = 0,82 J/g°<span>C.
</span>m(Fe) = 575 g.
ΔT₁ = 1181 - 1825 = -644°C.
Q₁ = -859306,5 J = -859,3 kJ.
Q₂ = m(Fe) · C · ΔT₂.
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7 0
3 years ago
Read 2 more answers
Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the
hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas.  So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

<u><em /></u>

Finally, the moles of gas used are 10.625 moles.

Learn more about molar mass:

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Answer: The given statement is TRUE.

Explanation:

An equilibrium reaction is one in which rate of forward reaction is equal to the rate of backward reaction.

Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.

For example for the given equilibrium reaction;

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K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

Thus the given statement that in calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the factors in the equilibrium expression is True.

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