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grin007 [14]
3 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]3 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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Answer:

c. ΔH° is positive and ΔS° is positive.

Explanation:

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In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

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At high temperatures phosphine (PH_3) dissociates into phosphorus and hydrogen by the following reaction: 4PH3 rightarrow P_4 +
Alex_Xolod [135]

Answer:a)p4=0.75mol

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P4=0.25×10^7mole

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So it means from the equation above

4 mol of PH3 will form 1mol of P4 and 6 mol of H2

Meaning that if 4mol of PH3 gives 1mol of P4

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In the same vein,3mol of PH3 will give 6×3/4 mol of H2= 4.5 Mol of H2

B) molar concentration =mole/volume

Meaning mole=molar concentration ×volume

Molar concentration is given as 5Km/m^3=5000mol/m^3

Volume=2L =2×1000=2000m^3

Nos of moles=5000×2000=10^7moles

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How many moles are in 0.0073 kg of tantalum?
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Tantalum is the 73rd element in the periodic table with an atomic mass equal to 180.95 g/mol. To determine the number of moles present in the given mass of tantalum above, we simply divide the mass by the atomic mass. 
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Therefore, there is approximately 0.0403 moles of tantalum in 0.0073 kg. 
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