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Oksana_A [137]
3 years ago
5

a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel

sius a sample of 2.35g of N2 g originally in a 25.0 L vessel at 20 degree Celsius is transferred to same 10.0 L vessel what is the partial pressure of so2 g in the larger container?what is the partial pressure of N2 g in this vessel ?what is the total pressure in the vessel?
Chemistry
1 answer:
eimsori [14]3 years ago
5 0

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

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where,            m = mass

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                   \Delta T = change in temperature

Putting the given values in the formula as follows.

       Enthalpy of neutralization = mS \Delta T

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                                                           = 14333.73 J

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Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.

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