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grin007 [14]
2 years ago
15

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

centration of B (M) Observed Initial Rate (M/s) 1 3.40 4.16 1.82 ✕ 10−4 2 4.59 4.16 3.32 ✕ 10−4 3 3.40 5.46 1.82 ✕ 10−4 Calculate the rate constant for this reaction.
Chemistry
1 answer:
Afina-wow [57]2 years ago
6 0

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

A\rightarrow B

Expt    [A] M        [B] M         Rate [M/s]

1          3.40         4.16           1.82*10^-4

2         4.59         4.16           3.32*10^-4

3.        3.40          5.46          1.82*10^-4

Rate = k[A]^{x}[B]^{y}

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2

To find y:

Divide rate of expt 3 by expt 1

\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0

Therefore: x = 2, y = 0

Rate = k[A]^{2}[B]^{0}

To find k

Use rate for expt 1:

k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1

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MA_775_DIABLO [31]

Answer:

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Explanation:

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HCl + KOH —> KCl + H2O

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From the balanced equation above, 1 mole of HCl produced 1 mole of KCl.

Therefore, 5.59x10^-3 moles of HCl will also produce 5.59x10^-3 moles of KCl.

From the illustration made above, we can see evidently that 5.59x10^-3 moles of HCl is required to produce 5.59x10^-3 moles of KCl

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2 years ago
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velikii [3]
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7 0
3 years ago
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marishachu [46]

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ipn [44]

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Good luck :)

5 0
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