Chemistry, 16.02.2021 14:00 richard80

Which substance below would have particles with the SLOWEST average velocity at 300 K? Fluorine, F2
Chlorine, Cl2
Methane, CH4
Xenon, Xe

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Answer from: jos0305

Likely Xenon, $\rm Xe$ .

Explanation:

The average kinetic energy of particles of a gas at a given temperature and pressure should be the same (regardless of the exact composition of the particle.)

For example, let m_1 and m_2 denote the mass of each particle of two gases, and let v_1 and v_2 denote the average speed of particles of each gas.

Average kinetic energy of particles of the first gas: $\displaystyle {\rm KE_1} = \frac{1}{2}\, m_1 \cdot ({v_1})^{2}$ .

Average kinetic energy of particles of the second gas: $\displaystyle {\rm KE_2} = \frac{1}{2}\, m_2 \cdot ({v_2})^{2}$ .

If the temperature of these two gases are the same, $\rm KE_1 = KE_2$ . Hence:

$\displaystyle \frac{1}{2}\, m_1 \cdot ({v_1})^{2}= \frac{1}{2}\, m_2 \cdot ({v_2})^{2}$ .

$\displaystyle \frac{v_1}{v_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}}$ .

In other words, the average velocity of particles of each gas (at the same temperature and pressure) is inversely proportional to the square root of the mass of one particle of each gas.

Similarly, at the same temperature and pressure, particles of the four gases in this question would have the same average kinetic energy. However, the gas with the largest per-particle mass would have the slowest average velocity for its particles.

Look up the relative atomic mass of the particles:

$\rm F$ :

. $\rm C$ : 12.011

. $\rm H$ : 1.008

. $\rm Cl$ : 35.45

. $\rm Xe$ : 131.293

Calculate the formula mass of each gas. That value would be proportional to the average mass of each particle of the gas:

$\begin{aligned}M(\rm F_2) &= (2 \times 18.998)\; \rm g\cdot mol^{-1}\\ &= 37.996\; \rm g\cdot mol^{-1}\end{aligned}$ . $\begin{aligned}M(\rm Cl_2) &= (2 \times 35.45)\; \rm g\cdot mol^{-1}\\ &= 70.90\; \rm g\cdot mol^{-1}\end{aligned}$ . $\begin{aligned}M(\rm CH_4) &= (12.011 + 4 \times 1.008)\; \rm g\cdot mol^{-1}\\ &= 16.043\; \rm g\cdot mol^{-1}\end{aligned}$ . $\begin{aligned}M(\rm Xe) &= 131.293\; \rm g \cdot mol^{-1}\end{aligned}$ .

Among these four gases, each particle of xenon, $\rm Xe$ , has the largest per-particle mass. Hence, at the same temperature and pressure, particles of Xenon would have the slowest average velocity among these gases.

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 :

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

Answer from: Quest

c-14 is carbon-14, or radiocarbon. it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.

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Which substance below would have particles with the SLOWEST average velocity at 300 K? Fluorine, F2...

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