Chemistry, 07.03.2020 05:09 kyllow5644

Use the following data to estimate the normal boiling point and the molar enthalpy of vaporization of lead: T (K) 1500 1600 1700 1800 1900 P (Torr) 19.72 48.48 107.2 217.7 408.2

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Answer from: arieannaensley0616

ΔHvap = 1.796 x 10⁵ J = 179.6 kJ

Tb = 1737 ºC

Explanation:

The expression we should use to solve this equation is given by

ln p = - ΔHvap/R x (1/T) +c

where ln p is the natural log of the pressues, ΔHvap is the enthalpy of vaporization, R is the gas constant (8.314 J/Kmol), T is the temperature and c is a constant. This is called the Clausius-Clayperon equation for the temperature dependence of pressure.

Note this equation has a form

y= mx + b

where m the slope is equal to - ΔHvap/R.

We have the values of T and P, therefore, we can determine the slope of these data to answer our question.

T(K) 1/T (K⁻¹) P (torr) lnp

1500 6.67 x 10⁻⁴ 19.72 2.98

1600 6.25 x 10⁻⁴ 48.48 3.88

1700 5.88 x 10⁻⁴ 107.2 4.67

1800 5.56 x 10⁻⁴ 217.7 5.38

1900 5.26 x 10⁻⁴ 408.2 6.01

Using a comercial software or calculating by hand the best slope, one is able to determine the desired values. In excel we get the equation of line:

y = -21598x + 17.38

with R²= 1

Therefore

- ( 21598 ) = -ΔHvap/ 8.314

⇒ ΔHvap = 1.796 x 10⁵ J = 179.6 kJ

and to calculate the boiling point we will use the equation for the best line just determined with p = 760 torr

ln p = -21598x + 17.38

where p is the pressure , and Tb is the normal boiling point:

ln ( 760 ) = -21598 ( 1/Tb ) + 17.38

Tb = 2009 K = 1737 ºC

Answer from: Quest

both adhesion and cohesion are attractions.

they are different in that adhesion is the force of attraction between the molecules of two different substances, whereas cohesion is the force of attraction between molecules of the same substance.

possible examples include capillary action, in which the adhesion between water and glass causes the water to rise in a narrow tube against the force of gravity.

because of hydrogen bonding, water is extremely cohesive.

this causes molecules on the surface of water to be drawn inward, which is why drops of water form beads on smooth surfaces.

hope this .

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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