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Lostsunrise [7]
3 years ago
9

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

A and 0.90 M of reagents B and C?
Trial [A]

(M) [B]

(M) [C]

(M) Initial rate

(M/s)

1 0.20 0.20 0.20 6.0×10−5

2 0.20 0.20 0.60 1.8×10−4

3 0.40 0.20 0.20 2.4×10−4

4 0.40 0.40 0.20 2.4×10−4
Chemistry
1 answer:
sladkih [1.3K]3 years ago
7 0

Answer:

\large\boxed{\large\boxed{0.0014M/s}}

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial  [A] (M)    [B] (M)   [C] (M)    Initial rate (M/s)

 1        0.20      0.20       0.20         6.0×10⁻⁵

2        0.20      0.20       0.60         1.8×10⁻⁴

3        0.40      0.20        0.20        2.4×10⁻⁴

4        0.40      0.40        0.20        2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

  • 0.60 / 0.2 = 3, and
  • 1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

  • 0.40/0.2 = 2
  • 2.4×10⁻⁴ /  6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

              a=2\\\\b=0\\\\c=1

Hence the rate is:

            rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

            6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}

<u>5. Rate law:</u>

       rate=k[A]^2C=0.0075[A]^2[C]

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

        rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]

        r=0.00136688M/s\approx 0.0014M/s

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Explanation:

Reaction is as follows:

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molecular mass of NH_3 is 17 g/mol

No. of mol of NH_3 = 11.5/17 = 0.676 mol

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No. of  mol of CH_4 = 10.5/16 = 0.656 mol

from the balanced chemical reaction, it is clear that 2-moles ammonia reacts with 3 moles oxygen and 2 moles methane to form 2 moles of HCN.

or, 1-mol ammonia reacts with 1.5 mol oxygen and 1 mol methane to form 1 mol of HCN.

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From the chemical equation,

3 moles oxygen produces 2 moles HCN

or one mole oxygen produces (2/3) moles HCN

0.375 moles oxygen produces (2/3) × 0.375 HCN = 0.25 moles of HCN  

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8 0
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4.5g * (1mol / 150g) = 0.03 moles

<em>Moles B:</em>

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0.03 moles A * (1 mole B / 3 moles A) = 0.01 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 3 moles A) = 0.01 moles of C.

0.01 moles of C * (200g / mol) = 2g are produced.

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2g / 2g * 100 = 100%

b. For a complete reaction of 0.03 moles of A are needed:

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As there are 0.04 moles of B, B is limiting reactant.

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0.03 moles A * (1 mole B / 1 moles A) = 0.03 moles of B

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0.03 moles A * (1 mole C / 1 moles A) = 0.03 moles of C.

0.03 moles of C * (200g / mol) = 6g are produced.

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6g / 2g * 100 = 300%

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4 0
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3.
pav-90 [236]

Answer:

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Explanation:

Hello there!

In this case, according to the given atoms of calcium, it is possible to calculate the mass of this element by considering that the definition of mole is in terms of atoms and also the atomic mass of calcium:

1molCa=6.022x10^{23}atoms Ca=40.08gCa

In such a way, by considering the following setup, we can obtain:

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Best regards!

8 0
3 years ago
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