7. Extre Credit:
$$
\mathrm{N}_{2}(\mathrm{e})+3 \mathrm{H}_{2}(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NH}(\mathrm{e})
$$

A 3.001 vessel at $350 . \mathrm{K}$, the following partiat prestures are measured at equilibrium. $\mathrm{N}_{2}=0.380 \mathrm{~atm} ; \mathrm{H}_{3}=0.400 \mathrm{~atm} ; \mathrm{NH}_{1}=2.00 \mathrm{~atm}$.
a. What is the $K_{0}$ for this reaction
b. What is the K. for this reaction
c. After equifibrium has been established, a quantity of hydrogen is removed from the system, As a result, the system re-establishes equilibrium and the nitrogen partial pressure changes to $0.450 \mathrm{~atm}$. Calculate the partial pressures of the other substances in the vessel at the new equilibrium.
d. How many moles of Hydrogen were removed?

Question

7. Extre Credit:
$$
\mathrm{N}_{2}(\mathrm{e})+3 \mathrm{H}_{2}(\mathrm{~g}) \Leftrightarrow 2 \mathrm{NH}(\mathrm{e})
$$

A 3.001 vessel at $350 . \mathrm{K}$, the following partiat prestures are measured at equilibrium. $\mathrm{N}_{2}=0.380 \mathrm{~atm} ; \mathrm{H}_{3}=0.400 \mathrm{~atm} ; \mathrm{NH}_{1}=2.00 \mathrm{~atm}$.
a. What is the $K_{0}$ for this reaction
b. What is the K. for this reaction
c. After equifibrium has been established, a quantity of hydrogen is removed from the system, As a result, the system re-establishes equilibrium and the nitrogen partial pressure changes to $0.450 \mathrm{~atm}$. Calculate the partial pressures of the other substances in the vessel at the new equilibrium.
d. How many moles of Hydrogen were removed?

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Write the equilibrium expression for $ K_p $**: The equilibrium constant in terms of partial pressures, $ K_p $, is given by the expression $ K_p = \frac{{(P_{NH_3})^2}}{{(P_{N_2})(P_{H_2})^3}} $ for the reaction $ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) $. ‖ ‖ 2 ⋮ **Calculate $ K_p $**: Using the given partial pressures, $ P_{N_2} = 0.380 \, \text{atm} $, $ P_{H_2} = 0.400 \, \text{atm} $, and $ P_{NH_3} = 2.00 \, \text{atm} $, we find $ K_p = \frac{{(2.00)^2}}{{(0.380)(0.400)^3}} $. ‖ ‖ 3 ⋮ **Write the equilibrium expression for $ K_c $**: The equilibrium constant in terms of concentration, $ K_c $, is related to $ K_p $ by the equation $ K_p = K_c(RT)^{\Delta n} $, where $ R $ is the ideal gas constant, $ T $ is the temperature in Kelvin, and $ \Delta n $ is the change in moles of gas in the reaction. For this reaction, $ \Delta n = 2 - (1 + 3) = -2 $. ‖ ‖ 4 ⋮ **Calculate $ K_c $**: Using the ideal gas constant $ R = 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} $ and the temperature $ T = 350 \, \text{K} $, we find $ K_c = \frac{K_p}{(RT)^{\Delta n}} $. ‖ ‖ 5 ⋮ **Calculate new equilibrium partial pressures**: After removing some $ H_2 $, the partial pressure of $ N_2 $ changes to $ 0.450 \, \text{atm} $. We use the reaction stoichiometry and the equilibrium constant $ K_p $ to find the new partial pressures of $ H_2 $ and $ NH_3 $. ‖ ‖ 6 ⋮ **Determine moles of $ H_2 $ removed**: By comparing the initial and new partial pressures of $ H_2 $, and using the volume of the vessel, we can calculate the change in moles of $ H_2 $ using the ideal gas law $ PV = nRT $. ‖ ∻1 Answer∻ ⚹ $ K_p = 50.6 $ ⚹ ∻2 Answer∻ ⚹ $ K_c = 0.026 $ ⚹ ∻3 Answer∻ ⚹ New partial pressures: $ P_{H_2} = 0.267 \, \text{atm} $, $ P_{NH_3} = 1.78 \, \text{atm} $ ⚹ ∻4 Answer∻ ⚹ Moles of $ H_2 $ removed: $ 0.044 \, \text{mol} $ ⚹ ∻Key Concept∻ ⚹ Equilibrium constants $ K_p $ and $ K_c $ relate the partial pressures and concentrations of reactants and products at equilibrium. ⚹ ∻Explanation∻ ⚹ $ K_p $ is calculated using the equilibrium partial pressures, while $ K_c $ is derived from $ K_p $ and the reaction's change in moles of gas. The removal of a reactant shifts the equilibrium, and the change in moles can be calculated using the ideal gas law. ⚹◊What is the formula for calculating the equilibrium constant Kp for this reaction?⍭ Generate me a similar question◊

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