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saw5 [17]
3 years ago
10

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m

ultiple compounds.
a. Wide melting point range __________
b. Experimental melting point is close to literature value _________
c. Experimental melting point is below literature value _________
d. Narrow melting point range __________
Chemistry
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

Wide melting point range - impure sample with multiple compounds

Experimental melting point is close to literature value - pure sample of a single compound

Experimental melting point is below literature value - impure sample with multiple compounds

Narrow melting point range - pure sample of a single compound

Explanation:

The melting point of substances are easily obtainable from literature such as the CRC Handbook of Physics and Chemistry.

A single pure substance is always observed to melt within a narrow temperature range. This melting temperature is always very close to the melting point recorded in literature for the pure compound.

However, an impure sample with multiple compounds will melt over a wide temperature range. We also have to recall that impurities lower the melting point of a pure substance. Hence, the experimental melting point of an impure sample with multiple compounds is always below the literature value.

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The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
9966 [12]

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The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0
2 years ago
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