Chemistry, 10.07.2019 12:30 cxmochaa

Atwenty-eight-liter volume of gas contains 11 g methane, 1.5-gram nitrogen and 16-gram carbon dioxide. determine partial pressure of each gas.

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Answer from: goldenhawk239

Based on Dalton's Law, for a mixture of gases, the total pressure is the sum of the partial pressure of each gas.

Partial pressure (p) of each gas is related to the total pressure (P) as follows:

p = X * P(1)

where X is the mole fraction of that gas

X = moles of a particular gas/total number of moles of all gases in the mixture(2)

Step 1: Calculate the moles of each gas

Mass of methane, CH4 = 11 g

Mass of nitrogen, N2 = 1.4 g

Mass of carbon dioxide, CO2 = 16 g

# moles of CH4 = 11 g/16 gmol-1 = 0.6875

# moles of N2 = 1.4/28 = 0.05

# moles of CO2 = 16/44 = 0.3636

Total moles = 0.6875+0.05+0.3636 = 1.1011

Step2: Calculate mole fractions of each gas

Based on equation (2)

X(CH4) = 0.6875/1.1011 = 0.6244

X(N2) = 0.05/1.1011 = 0.0454

X(CO2) = 0.3636/1.1011 = 0.3302

Step 3: Calculate the total pressure

Based on ideal gas equation:

PV = nRT

given that;

V = 28 L

n = total moles = 1.1011

R = gas constant = 0.0821 Latm/mol-K

Since temp T is not given, let us consider room temperature of 25 C = 25 + 273 = 298 K

Now, P = nRT/V = 1.011*0.0821*298/28 = 0.962 atm

Step 3: Calculate partial pressures

Based on equation:

p(CH4) = 0.6244*0.962 atm = 0.601 atm

P(N2) = 0.0454*0.962 atm = 0.044 atm

P(CO2) = 0.3302*0.962 atm = 0.318 atm

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

hurricanes form when warm, moist air from the ocean surface quickly rises into the atmosphere and forms storm clouds and rain. when the clouds are formed, heat is released which warms the cool air in the upper regions of the atmosphere so that the cycle can continue. which two phases of the water cycle are repeated in this process?

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Atwenty-eight-liter volume of gas contains 11 g methane, 1.5-gram nitrogen and 16-gram carbon dioxid...