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At\[500K\], the equilibrium constant for reaction \[cis - {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}_{(g)} \Leftrightarrow trans{\text{ - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}_{(g)}\] is 0.6. At the same temperature, the equilibrium constant $trans - {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}_{(g)} \Leftrightarrow {\text{cis - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}_{(g)}$will be
A.1.67
B.0.6
C.1.76
D.1.64

Answer Verified Verified
Hint: We must know that the equilibrium constant of the reverse reaction can calculate by reciprocal the equilibrium constant of the original reaction (forward reaction).
Formula used:
We can use the formula to calculate the equilibrium constant for the second reaction is,
${{\text{K}}^{\text{'}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{K}}}$
The equilibrium concentration of the second reaction (reverse reaction) is ${{\text{K}}^{\text{'}}}.$
The equilibrium concentration of the first reaction (forward reaction) is ${\text{K}}.$

Complete step by step answer:
The first (forward) reaction is,
${\text{cis - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \Leftrightarrow {\text{trans - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$
The value of equilibrium constant for the above reaction is ${\text{0}}{\text{.6}}{\text{.}}$
The reaction ${\text{cis - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \Leftrightarrow {\text{trans - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is reversed to get reaction,
$trans - {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \Leftrightarrow {\text{cis - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}.$
For the second (reversed) reaction, the equilibrium constant value is calculated by taking the reciprocal of the equilibrium constant of the first reaction.
Hence, the value of the equilibrium constant for the second (reverse) reaction is the reciprocal of the value of the value of the equilibrium constant for the first (forward) reaction.
The equilibrium constant for the second reaction is,
$
  {{\text{K}}^{\text{'}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{K}}} \\
  {{\text{K}}^{\text{'}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{0}}{\text{.6}}}} \\
  {{\text{K}}^{\text{'}}}{\text{ = 1}}{\text{.66}} \\
  {{\text{K}}^{\text{'}}}{\text{ = 1}}{\text{.67}} \\
 $
The equilibrium constant for the reaction trans-${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \Leftrightarrow {\text{cis - }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is $1.67.$
Option (a) is correct.

Note:
We must know that in a chemical reaction, chemical equilibrium is the state in which the rates of forward reaction and the reverse reaction are equal. The result is that the concentration of the reactants and the products remains unchanged.
Equilibrium expression for the reverse reaction is reciprocal of the forward reaction.
When the coefficients in a balanced equation are multiplied with a faction n, the equilibrium expression is raised to the ${{\text{n}}^{{\text{th}}}}$ power.