Question

give the expression for $k_{eq}$ for the reverse reaction. all are gases in a closed container. $\ce{c2h6 + 2 cl2 <=> c2h4cl2 + 2 hcl}$ the concentrations for both $\ce{cl2}$ and $\ce{hcl}$ are squared due to their coefficients. $k_{eq}^r = \frac{\\_\\_1\\_\\_\\_\\_2\\_\\_^2}{\\_\\_3\\_\\_\\_\\_4\\_\\_^2}$ a. $\ce{h3o^{+1}}$ b. $\ce{oh^{-1}}$ c. $\ce{c2h6}$ d. $\ce{c2h4cl2}$ e. $\ce{cl2}$ f. $\ce{hcl}$ g. $\ce{h2so3}$ h. $\ce{hc2h3o2}$ i. $\ce{c2h3o2^{-1}}$ j. $\ce{h2s}$ k. $\ce{hs^{-1}}$ l. $\ce{s^{-2}}$ m. $\ce{hx}$ n. $\ce{x^{-1}}$ o. 1 p. 2 q. 3 r. 0.5000 s. $5.5 \times 10^{-5}$ t. $3.5 \times 10^{-4}$ u. $8.00 \times 10^{-5}$ v. $1.75 \times 10^{-6}$ w. $1.11 \times 10^{-8}$ x. $5.5 \times 10^{-2}$ y. 0.7000 z. $1.60 \times 10^{-5}$ aa. $1.143 \times 10^{-5}$ bb. 0.4900 cc. $1.104 \times 10^{-5}$ dd. 0.7100 ee. $1.26 \times 10^{-5}$ ff. 4.901 gg. 7.93 hh. 4.942 ii. 4.957

Explanation:

Step1: Identify Reverse Reaction

The original reaction is \( \ce{C2H6 + 2 Cl2 <=> C2H4Cl2 + 2 HCl} \). The reverse reaction will be \( \ce{C2H4Cl2 + 2 HCl <=> C2H6 + 2 Cl2} \).

Step2: Recall \( K_{eq} \) for Reverse Reaction

For a reaction \( aA + bB <=> cC + dD \), \( K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b} \). For the reverse reaction, \( K_{eq}^r = \frac{[A]^a[B]^b}{[C]^c[D]^d} \).

Step3: Apply to Reverse Reaction

For reverse reaction \( \ce{C2H4Cl2 + 2 HCl <=> C2H6 + 2 Cl2} \), the \( K_{eq}^r \) expression is \( \frac{[\ce{C2H6}][\ce{Cl2}]^2}{[\ce{C2H4Cl2}][\ce{HCl}]^2} \). So:

• 1: \( \ce{C2H6} \) (Option C)
• 2: \( \ce{Cl2} \) (Option E)
• 3: \( \ce{C2H4Cl2} \) (Option D)
• 4: \( \ce{HCl} \) (Option F)

Answer:

1: C. \( \ce{C2H6} \)
2: E. \( \ce{Cl2} \)
3: D. \( \ce{C2H4Cl2} \)
4: F. \( \ce{HCl} \)
\( K_{eq}^r = \frac{[\ce{C2H6}][\ce{Cl2}]^2}{[\ce{C2H4Cl2}][\ce{HCl}]^2} \)