Assuming the temperature and volume remain constant, changes to the pressure in

Question

Assuming the temperature and volume remain constant, changes to the pressure in the reaction vessel will directly correspond to changes in the number of moles based on the ideal gas law.

Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in part A, and the temperature and volume were constant at values of 303 K and 2.00 L, respectively. If the pressure was 11.3 atm prior to the reaction, what would be the expected pressure after the reaction was completed?

Express the pressure in atmospheres to three significant digits.

Part A First, let us consider the reaction between nitrogen (N2) and hydrogen (H2) gases to form ammonia gas (NHs). In the Simulation, you can run the reaction between nitrogen and hydrogen by selecting the Experiment tab, clicking the Run Experiment button, and selecting Nitrogen and Hydrogen from the dropdown menu. You can choose any starting amounts you wish, but compare the results when you use (1) stoichiometric amounts of each reactant to (2) one of the reactants is in excess to familiarize yourself with how the Simulation If you started with 0.227 mol of N2 and 0.681 mol of H2, and they completely reacted in the reaction vessel, determine total moles of gas particles (n) there are during the initial and final conditions. Additionally, determine the ratio of the number of gas particles in the products to that of the reactants, then complete the statements below. Match the values in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer Hints Reset Help nitially, there are 0.908 mol of gas particles (Minit). After the reaction is completed, there are 0.454 mol of gas particles (nfin) Therefore, the ratio of product to reactant particles ( ) is | 0.5 0.3 2.0 0.227 mol 0.681 mol 3.0 Submit My Answers Give Up Correct Now that you have established the relationship between the number of gas particles before and after a reaction between stoichiometric amounts of nitrogen and hydrogen gases, let us examine how this affects the pressure in the reaction vessel.

Explanation / Answer

N2(g) + 3H2(g) ------> 2NH3(g)

Initial Number of moles = moles of N2 + moles of H2 = 0.227 + 0.681 = 0.908 moles

One mole of N2 will react with 3 moles of H2 to form 2 moles of NH3

Moles of NH3 formed = 2 * moles of N2 = 2 * 0.227 = 0.454 moles

Ration of Product to reactant particles = 0.454/0.908 = 0.5 moles

b)

Using the ideal gas equation

PV = nRT

Since the temperature and volume were kept constant, so we can write

P(in)/n(in) = P(final)/n(final)

11.3/0.908 = P(final)/0.454

P(final) = 11.3 * 0.454/0.908 = 5.65 atm

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