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vodomira [7]
3 years ago
6

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

Chemistry
1 answer:
algol [13]3 years ago
7 0

Answer:

  • 820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

  • KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

  • First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

  • Second, convert 0°C to kelvin:

         T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

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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
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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:

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  • 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
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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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