For the reaction 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), K_c = 8.0 at a certain temperature. What concentration of NOCl must be put into an empty 4.00 L reaction vessel in order that the equilibrium concentration of NOCl be 1.00 M

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Given the equilibrium constant $K_c = 8.0$ for the reaction $2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$, and the desired equilibrium concentration of NOCl is 1.00 M, we can set up an ICE (Initial, Change, Equilibrium) table to solve for the initial concentration of NOCl needed.At equilibrium, the concentration of NOCl decreases by $x$ to reach 1.00 M, so the change for NOCl is $-x$. Since the reaction consumes 2 moles of NOCl for every 2 moles of NO and 1 mole of $Cl_2$ produced, the changes for NO and $Cl_2$ are $+x$ and $+0.5x$, respectively.The equilibrium expression is:$$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = 8.0$$Substituting the equilibrium concentrations gives:$$8.0 = \frac{(1+x)^2(0.5x)}{(1-x)^2}$$Solving this equation directly for $x$ is complex without knowing the initial concentration, but we know at equilibrium:- $[NOCl] = 1.00 M$- $K_c = 8.0$We can simplify the problem by understanding that the initial concentration of NOCl ($C_i$) minus the change ($x$) gives the equilibrium concentration (1.00 M). The amount of NOCl that reacts ($x$) will be twice the concentration of $Cl_2$ produced because for every 2 moles of NOCl that react, 1 mole of $Cl_2$ is produced.Given the stoichiometry of the reaction, if the equilibrium concentration of $Cl_2$ is $y$, then the concentration of NO will also be $2y$ (because they are produced in a 1:2 ratio), and $x = 2y$.Using the equilibrium constant expression:$$8.0 = \frac{(2y)^2(y)}{(1.00)^2}$$Solving for $y$ gives the equilibrium concentrations of the products. However, the direct calculation from the given options is not straightforward without algebraic manipulation or trial and error with the options provided.The correct approach involves recognizing that the initial concentration must be higher than the equilibrium concentration of NOCl (1.00 M) and that the equilibrium setup allows for the calculation of how much NOCl is consumed to reach equilibrium. The initial concentration minus the amount consumed equals the equilibrium concentration.Given the complexity of solving for $x$ directly from the equilibrium expression and the initial conditions, the correct answer involves understanding the relationship between the initial and equilibrium concentrations and the stoichiometry of the reaction. The initial concentration must account for the NOCl consumed (to form NO and $Cl_2$) and still have 1.00 M NOCl at equilibrium. The correct answer (D) 3.52 M, is found by recognizing it as the only option that, when the appropriate amount of NOCl is consumed to reach the equilibrium concentrations of products and remaining reactants, satisfies the given equilibrium constant and conditions.

For the Reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g), Kc =

For the Reaction 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), K_c =