Given the equation delta G = -RTln(Keq), plugging in the correct values for the variables will lead us to the correct answer. Using e = 2.7, R = 8.314 J/molK, T= 298 K, and G = -4955 J, we can calculate that K = 7.4.

When K is greater than 1, the products are favored and the reaction proceeds forward spontaneously. This must be true of the reaction, since it spontaneously occurs when the metal comes into contact with the water.

For any equilibrium expression, solids and liquids are excluded because their concentrations are presumed to be constant during the reaction. The equilibrium expression is defined as the ratio of the concentration of products divided by the concentration of the reactants. Each reactant or product is raised to the power corresponding to its coefficient in the balanced chemical equation.

Using an arbitrary example:

Compare this to our question. Remember to exclude the solid magnesium!

The equilibrium constant, , is the ratio of the concentration of products raised to their coefficients, over the concentration of reactants raised to their coefficients. To find its value, we first need to balance the equation and then consider only the products and reactants that actually have concentrations (i.e. aqueous and gaseous species). Liquids and solids can be omitted from the calculation.

We note that chlorine only appears once on the right, so we add a 2 coefficient to balance the equation: 

Since is a solid (most hydroxides precipitate, unless they are paired with alkali metals, barium, or calcium) it will not be included in the equilibrium constant calculation.

The chloride ion concentration is raised to the second power because of the 2 coefficient that we added. So, putting products (species on the right side, other than the solid) on the top and reactants (species on the left) on bottom, we get: 

We can set up the equilibrium expression by placing the products, raised to the power of their coefficients, in the numerator and reactants, raised to the power of their coefficients, in the denominator:

We are given the value of the equilibrium constant and the equilibrium concentration of . Using stoichiometric coefficients, we can determine that a variable concentration, , of  will be present at equilibrium, and  of  will be present at equilibrium. Plugging in:

Simplify:

Now we can solve for  (which is the equilibrium concentration of , as we assigned) by dividing both sides by 4 and then taking the cube root.

Rounding to 2 sig figs (to match the two numbers we were given by the problem), we get:

I) Decreasing the temperature would take away heat from the system (a product), driving the reaction towards the products. II) Evaporating product would take a product away from the system, driving the reaction towards the products. III) Adding a catalyst only affects the rate of the reaction and does not effect equilibrium.

According to Le Chatelier's principle, when you compress a system, its volume decreases, so partial pressure of the all the gases in the system increases. The system will act to try to decrease the pressure by decreasing the moles of gas. 

In an endothermic reaction, heat can be treated as a reactant. Thus, if you add more reactant (heat), the system will shift to get rid of the extra reactant and shift to the right to form more products.

In an exothermic reaction, heat can be treated as a product. Thus, if you add more product (heat), the reaction will shift to the left to form more reactants. 

By increasing the concentration of one of the reactants, the reaction will compensate by shifting to the right to increase production of products.

Increasing the concentration of one of the products (such as increasing [C]), however, would have the opposite effect. Increasing the temperature of an exothermic reaction would shift the reaction to the left, while increasing the temperature of an endothermic reaction would lead to a rightward shift. Finally, decreasing the volume leads to an increase in partial pressure of each gas, which the system compensates for by shifting to the side with fewer moles of gas. In this case, the right side has three moles of gas, while the left side has two; thus decreasing volume would shift equilibrium to the left.