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]
