Question

scoring: your score will be based on the number of correct matches minus the number of incorrect matches. there is no penalty for missing matches.
use the references to access important values if needed for this question.
the equilibrium constant, ( k_c ), for the following reaction is 10.5 at 350 k.
( 2ch_2cl_2(g)
ightleftharpoons ch_4(g)+ccl_4(g) )
assuming that you start with only ( ch_2cl_2 ), describe the relative abundance of each species present at equilibrium.
ch₄
ccl₄
ch₂cl₂
lower
higher
cant tell
clear all

Explanation:

Step1: Write the equilibrium constant expression

For the reaction \(2CH_{2}Cl_{2}(g)
ightleftharpoons CH_{4}(g)+CCl_{4}(g)\), the equilibrium constant \(K_{c}=\frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}}\)

Step2: Analyze the value of \(K_{c}\)

Given \(K_{c} = 10.5\) (a value greater than \(1\)). When \(K_{c}>1\), the numerator \([CH_{4}][CCl_{4}]\) is greater than the denominator \([CH_{2}Cl_{2}]^{2}\) at equilibrium.

Since the stoichiometric coefficients of \(CH_{4}\) and \(CCl_{4}\) are \(1\) (equal), and we start with only \(CH_{2}Cl_{2}\), at equilibrium \([CH_{4}]=[CCl_{4}]\).

Let \([CH_{4}]=[CCl_{4}]=x\) and \([CH_{2}Cl_{2}]=y\). Then \(K_{c}=\frac{x\times x}{y^{2}}=\frac{x^{2}}{y^{2}} = 10.5\), so \(x>y\)

Answer:

\([CH_{4}]\): Higher, \([CCl_{4}]\): Higher, \([CH_{2}Cl_{2}]\): Lower