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The combustion of ethylene proceeds by the reaction

[tex]\[ C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) \][/tex]

When the rate of disappearance of [tex]\( O_2 \)[/tex] is [tex]\( 0.23 \, M \, s^{-1} \)[/tex], the rate of disappearance of [tex]\( C_2H_4 \)[/tex] is

A. 0.077 [tex]\( M \, s^{-1} \)[/tex]
B. 0.15 [tex]\( M \, s^{-1} \)[/tex]
C. 0.46 [tex]\( M \, s^{-1} \)[/tex]
D. 0.69 [tex]\( M \, s^{-1} \)[/tex]
E. 0.35 [tex]\( M \, s^{-1} \)[/tex]

Sagot :

GinnyAnsweridn
To solve for the rate of disappearance of [tex]\( \text{C}_2\text{H}_4 \)[/tex] when given the rate of disappearance of [tex]\( \text{O}_2 \)[/tex], we can utilize the stoichiometry of the chemical reaction involved:

[tex]\[ \text{C}_2\text{H}_4(g) + 3 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]

Here we can see from the balanced chemical equation that 1 mole of [tex]\( \text{C}_2\text{H}_4 \)[/tex] reacts with 3 moles of [tex]\( \text{O}_2 \)[/tex]. This means that for every mole of [tex]\( \text{C}_2\text{H}_4 \)[/tex] consumed, 3 moles of [tex]\( \text{O}_2 \)[/tex] are consumed.

Given:
- Rate of disappearance of [tex]\( \text{O}_2 \)[/tex] = [tex]\( 0.23 \, \text{M s}^{-1} \)[/tex]

We need to find the rate of disappearance of [tex]\( \text{C}_2\text{H}_4 \)[/tex].

Using stoichiometry, we know that the rate at which [tex]\( \text{C}_2\text{H}_4 \)[/tex] is disappearing is one-third the rate at which [tex]\( \text{O}_2 \)[/tex] is disappearing, because 1 mole of [tex]\( \text{C}_2\text{H}_4 \)[/tex] reacts with 3 moles of [tex]\( \text{O}_2 \)[/tex].

Thus:
[tex]\[ \text{Rate of disappearance of } \text{C}_2\text{H}_4 = \frac{1}{3} \times 0.23 \, \text{M s}^{-1} \][/tex]

Calculating this, we get:
[tex]\[ \text{Rate of disappearance of } \text{C}_2\text{H}_4 = 0.07666666666666666 \, \text{M s}^{-1} \][/tex]

This can be approximated to:
[tex]\[ \text{Rate of disappearance of } \text{C}_2\text{H}_4 \approx 0.077 \, \text{M s}^{-1} \][/tex]

From the given options, the rate of disappearance of [tex]\( \text{C}_2\text{H}_4 \)[/tex] is [tex]\( 0.077 \, \text{M s}^{-1} \)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{0.077} \][/tex]
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