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Bond [772]
3 years ago
6

d. Calculate the initial reaction rate that would be seen in a solution in which [BF3] is 5.0 x10 -1M and [NH3] is 2.0 x10 -2 M

Chemistry
1 answer:
Rashid [163]3 years ago
6 0

Answer:

r=0.0341\frac{M}{s}

Explanation:

Hello!

In this case, considering the given table, we are able to represent the symbolic rate law as shown below:

r=k[BF_3]^m[NH_3]^n

Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):

- Experiment 1 and 2 for the calculation of n:

\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}

So we plug in to obtain:

\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1

So the order of reaction with respect to NH3 is 1.

- Experiment 3 and 4 for the calculation of m

\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}

So we plug in to obtain:

\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1

So the order of reaction with respect to BF3 is also 1.

Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:

k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}

Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:

r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}

Best regards!

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Answer :  The heat of the reaction is -221.6 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter

q_{rxn}=-q_{cal}

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q_{rxn} = heat released by the reaction = ?

q_{cal} = heat absorbed by the calorimeter

c_{cal} = specific heat of calorimeter = 97.1kJ/^oC=97100J/^oC

\Delta T = change in temperature = (T_{final}-T_{initial})=(27.282-25.000)=2.282^oC

Now put all the given values in the above formula, we get:

q_{cal}=(97100J/^oC)\times (2.282^oC)

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Which atom’s ionization energy is greater than that of phosphorus (P)? A. Ba B. K C. As D. Cl
aalyn [17]
<h3>Answer:</h3>

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<h3>Explanation:</h3>

                           Ionization Energy is defined as, "the minimum energy required to knock out or remove the valence electron from valence shell of an atom".

Trends in Periodic table:

               <em>Along Periods:</em>

                                        Ionization Energy increases from left to right along the periods because moving from left to right in the same period the number of protons (atomic number) increases but the number of shells remain constant hence, resulting in strong nuclear interactions and electrons are more attracted to nucleus hence, requires more energy to knock them out.

               <em>Along Groups:</em>

                                        Ionization energy decreases from top to bottom along the groups because the number of shells increases and the distance between nucleus and valence electrons also increases along with increase in shielding effect provided by core electrons. Therefore, the valence electrons experience less nuclear attraction and are easily removed.

<h3>Conclusion:</h3>

                   As Barium is present down the group and to the left of periodic table, K present at the left of the periodic table, Arsenic present below Phosphorous will have less ionization energies as compared to P. Hence, only Chlorine present at the right extreme of periodic table and right to P will have greater Ionization energy.

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<h3>How to determine the density </h3>

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The answer to your question is below

Explanation:

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