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labwork [276]
3 years ago
9

Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:

Chemistry
1 answer:
Drupady [299]3 years ago
5 0

Answer:

r = k [A]^{2}[B]^{2}

Explanation:

A + B + C ⟶ D

\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}

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How old is a bone if it now .3125 of C-14 when it originally had 80.0g of C-14
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<h3>Answer:</h3>

42960 years

<h3>Explanation:</h3>

<u>We are given;</u>

  • Remaining mass of C-14 in a bone is 0.3125 g
  • Original mass of C-14 on the bone is 80.0 g
  • Half life of C-14 is 5370 years

We are required to determine the age of the bone;

  • Using the formula;
  • Remaining mass = Original mass × 0.5^n , where n is the number of half lives.

Therefore;

0.3125 g = 80.0 g × 0.5^n

3.90625 × 10^-3 = 0.5^n

  • Introducing logarithm on both sides;

log 3.90625 × 10^-3 = n log 0.5

Solving for n

n = log 3.90625 × 10^-3 ÷ log 0.5

   = 8

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Age of the rock = 5370 years × 8

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Thus, the bone is 42960 years old

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Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of
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Explanation:

1. Write down the balanced chemical reaction:

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- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

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For the oxygen:

81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

\frac{0.81}{2}=0.41

For the oxygen:

\frac{2.54}{19}=0.13

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

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