consider the following reversible reaction.\n$2h_2o(g) \\leftrightarrow 2h_2(g)+o_2(g)$\nwhat is the…

consider the following reversible reaction.\n$2h_2o(g) \\leftrightarrow 2h_2(g)+o_2(g)$\nwhat is the equilibrium constant expression for the given system?\n$k_{eq}=\\frac{h_2o}{h_2o_2}$\n$k_{eq}=\\frac{h_2o^2}{h_2^2o_2}$\n$k_{eq}=\\frac{h_2^2o_2}{h_2o}$\n$k_{eq}=\\frac{h_2^2o_2}{h_2o^2}$

consider the following reversible reaction.\n$2h_2o(g) \\leftrightarrow 2h_2(g)+o_2(g)$\nwhat is the equilibrium constant expression for the given system?\n$k_{eq}=\\frac{h_2o}{h_2o_2}$\n$k_{eq}=\\frac{h_2o^2}{h_2^2o_2}$\n$k_{eq}=\\frac{h_2^2o_2}{h_2o}$\n$k_{eq}=\\frac{h_2^2o_2}{h_2o^2}$

Answer

Explanation:

Step1: Recall $K_{eq}$ definition

For reaction $aA + bB \leftrightarrow cC + dD$, $K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Step2: Match reaction coefficients

Reaction: $2H_2O(g) \leftrightarrow 2H_2(g)+O_2(g)$, so $a=2, A=H_2O; c=2, C=H_2; d=1, D=O_2$

Step3: Plug into $K_{eq}$ formula

Substitute values: $K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$

Answer:

$K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}$ (the fourth option)