Producer gas is manufactured industrially by passing air over red-hot coke, yielding a mix… — JAMB UTME

Producer gas is manufactured industrially by passing air over red-hot coke, yielding a mixture consisting approximately of $1$ mole of $\text{CO}$ to $2$ moles of $\text{N}_2$ . Water gas is produced by passing steam over incandescent coke, yielding an equimolar mixture of $\text{CO}$ and $\text{H}_2$ . If equal volumes of producer gas and water gas are allowed to effuse through identical porous barriers under the same conditions of temperature and pressure, which of the following statements correctly compares their initial rates of effusion? $(\text{Atomic masses: } \text{H} = 1, \text{C} = 12, \text{N} = 14, \text{O} = 16)$

Water gas effuses faster than producer gas because its average molar mass ( $15\text{ g mol}^{-1}$ ) is less than that of producer gas ( $28\text{ g mol}^{-1}$ ).Cevap
B
Producer gas effuses faster than water gas because the rate of effusion is directly proportional to the average molar mass of the gaseous mixture.
C
Both gas mixtures effuse at the exact same rate because carbon(II) oxide and nitrogen gas are gaseous allotropes of carbon with identical molecular masses.
D
Water gas effuses at twice the rate of producer gas because one mole of any gas occupies $22.4\text{ dm}^3$ at STP while producer gas occupies $44.8\text{ dm}^3$ per mole of carbon.

Cevap

Water gas effuses faster than producer gas because its average molar mass (

15\text{ g mol}^{-1}

) is less than that of producer gas (

28\text{ g mol}^{-1}

Water gas consists of an equimolar mixture of carbon(II) oxide and hydrogen gas, giving an average molar mass of

\frac{28 + 2}{2} = 15\text{ g mol}^{-1}

. Producer gas consists of carbon(II) oxide and nitrogen gas in a

1:2

molar ratio, giving an average molar mass of

\frac{28 + 2(28)}{3} = 28\text{ g mol}^{-1}

. According to Graham's Law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass (

R \propto \frac{1}{\sqrt{M}}

). Because water gas has a significantly lower average molar mass (

15\text{ g mol}^{-1}

) than producer gas (

28\text{ g mol}^{-1}

), water gas effuses faster.

Adım Adım Çözüm

Calculate the average molar mass of producer gas

Molar mass of

\text{CO} = 12 + 16 = 28\text{ g mol}^{-1}

, Molar mass of

\text{N}_2 = 2 \times 14 = 28\text{ g mol}^{-1}

. For a

1:2

mole ratio of

\text{CO} : \text{N}_2

\bar{M}_{\text{producer}} = \frac{1(28) + 2(28)}{1 + 2} = 28\text{ g mol}^{-1}

To compare effusion rates, the average molar mass of the gas mixture must first be determined.

Calculate the average molar mass of water gas

Molar mass of

\text{CO} = 28\text{ g mol}^{-1}

, Molar mass of

\text{H}_2 = 2 \times 1 = 2\text{ g mol}^{-1}

. For an equimolar (

1:1

) mixture of

\text{CO} : \text{H}_2

\bar{M}_{\text{water gas}} = \frac{1(28) + 1(2)}{1 + 1} = 15\text{ g mol}^{-1}

The average molar mass of water gas is required to apply Graham's Law.

Apply Graham's Law of Effusion to compare rates

\frac{R_{\text{water gas}}}{R_{\text{producer gas}}} = \sqrt{\frac{\bar{M}_{\text{producer}}}{\bar{M}_{\text{water gas}}}} = \sqrt{\frac{28}{15}} \approx 1.37. Since

\bar{M}_{\text{water gas}} < \bar{M}_{\text{producer}}

, water gas effuses faster.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (

R \propto \frac{1}{\sqrt{M}}

Anahtar Kavram

Composition of industrial fuel gases (producer gas vs water gas) and application of Graham's Law of effusion to gas mixtures

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