Question #5 Achieve Week #1

[Diego Salgues 2K]

Consider the reaction of NH3 and I2 to give N2 and HI.
2NH3(g)+3I2(g)↽−−⇀N2(g)+6HI(g)K

Using two or more of the given equations, determine the equilibrium constant, K , for the reaction of NH3 with I2.

H2(g)+I2(g)↽−−⇀2HI(g)I2(g)↽−−⇀2I(g)N2(g)+3H2(g)↽−−⇀2NH3(g)H2(g)+Cl2(g)↽−−⇀2HCl(g)Ka=160Kb=2.1×10−3Kc=3.6×10−2Kd=4.0×1018

I have a question about knowing why Kc^3 of 2HI is equal to 6HI? Why doesn't it multiply by 3, and instead is brought to the power of 3?

[Aaron Li 1D]

If you were to rewrite the K equation for the equilibrium of the system, then each of the concentration values would be increased by the power of 3, making the overall K increase by the power of 3. The coefficients of the reaction are used as exponents in the K equation, so increasing the coefficients by a factor will result in an increase by an exponent of that factor.

[Adele Nguyen 2G]

In the K expression, the stoichiometric coefficients are used for the numbers that the concentrations/partial pressures of the products/reactants are placed to the power of. Similarly, since the first equation is multiplied by 3 (essentially the stoichiometric coefficients are now three times greater), the equilibrium constant is brought to the power of 3 to make the K expression still equal.

[Ethan Famas 1H]

Recall the formula for finding K: both the products and reactants are raised to the number power which matches their coefficients in the equation (refer to the picture). So, if you multiply an equation by three (multiplying each of the coefficients within the equation by 3), you multiply each of the powers in the K formula by 3. That is the same thing as cubing the former K constant.