Using Le Chatelier's equation, estimate the LFL of a mixture of 0.7 mole fraction CO and 0.3 mole fraction ethane (CO LFL 12.5%, ethane LFL 2.0%). The estimated LFL is closest to which value?

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Multiple Choice

Using Le Chatelier's equation, estimate the LFL of a mixture of 0.7 mole fraction CO and 0.3 mole fraction ethane (CO LFL 12.5%, ethane LFL 2.0%). The estimated LFL is closest to which value?

Explanation:
Le Chatelier’s mixing rule for flammable limits combines the components by weighting their reciprocals of LFLs with their mole fractions. The idea is 1/LFL_mix = sum(x_i / LFL_i), where x_i are the mole fractions and LFL_i are the individual lower flammable limits (as fractions or percentages consistently). Here, LFLs are CO 12.5% and ethane 2.0%; mole fractions are CO 0.7 and ethane 0.3. Using percent units: 0.7/12.5 = 0.056 and 0.3/2.0 = 0.15. Sum = 0.206. Invert to get LFL_mix = 1/0.206 ≈ 4.85% by volume. So the mixture’s LFL is about 4.85% vol, which is closest to 4.50% in the given options. The result drops toward the lower LFL component (ethane) because it has a much lower LFL, and it comprises a substantial fraction of the mixture.

Le Chatelier’s mixing rule for flammable limits combines the components by weighting their reciprocals of LFLs with their mole fractions. The idea is 1/LFL_mix = sum(x_i / LFL_i), where x_i are the mole fractions and LFL_i are the individual lower flammable limits (as fractions or percentages consistently).

Here, LFLs are CO 12.5% and ethane 2.0%; mole fractions are CO 0.7 and ethane 0.3. Using percent units: 0.7/12.5 = 0.056 and 0.3/2.0 = 0.15. Sum = 0.206. Invert to get LFL_mix = 1/0.206 ≈ 4.85% by volume.

So the mixture’s LFL is about 4.85% vol, which is closest to 4.50% in the given options. The result drops toward the lower LFL component (ethane) because it has a much lower LFL, and it comprises a substantial fraction of the mixture.

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