At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

When we multiply a chemical reaction by a factor, the relationship between the equilibrium constants is given by: \(K_{new} = K^x\), where \(x\) is the constant by which we scale the reaction. From this, we get: \[K_{a} = K^{\frac{1}{2}} = (1.3 \times 10^{-2})^{\frac{1}{2}}\] b. Reverse the reaction:

The equilibrium constant of the reverse reaction is given by \(K_{new} = \frac{1}{K}\), such that: \[K_{b} = \frac{1}{1.3 \times 10^{-2}}\] c. Combine reverse and half of the initial reaction:

As the new reaction is a combination of the half reaction and reverse reaction, we can multiply their \(K\) values to find the new \(K\). First, we find the half reaction \(K\): \(K_{half} = K^{\frac{1}{2}} = (1.3 \times 10^{-2})^{\frac{1}{2}}\), and then the new \(K\) equation: \[K_{c} = K_{b} \times K_{half} = \frac{1}{1.3 \times 10^{-2}} \times (1.3 \times 10^{-2})^{\frac{1}{2}}\] d. Multiply the given reaction by 2:

As discussed in part a, in scaled reactions, we have \(K_{new} = K^x\), and for two times the reaction, we get: \[K_{d} = K^2 = (1.3 \times 10^{-2})^2\] #Step 3: Calculate the equilibrium constants for each new reaction# Perform the calculations for each part: a. \(K_{a} = (1.3 \times 10^{-2})^{\frac{1}{2}} \approx 0.114\) b. \(K_{b} = \frac{1}{1.3 \times 10^{-2}} \approx 76.92\) c. \(K_{c} = \frac{(1.3 \times 10^{-2})^{\frac{1}{2}}}{1.3 \times 10^{-2}} \approx 8.79\) d. \(K_{d} = (1.3 \times 10^{-2})^2 \approx 1.69 \times 10^{-4}\) #Step 4: Present the equilibrium constants for each new reaction# The equilibrium constants for the new chemical reactions are: a. \(K_{a} \approx 0.114\) b. \(K_{b} \approx 76.92\) c. \(K_{c} \approx 8.79\) d. \(K_{d} \approx 1.69 \times 10^{-4}\)