A chemist mixes together 0.450 mol N2 with 0.450 mol H2 in a sealed container. The mixture is heated and allowed to reach equilibrium. At equilibrium, the mixture contains 0.400 mol N2 and the total pressure is 500 kPa. Calculate Kp.

balanced equation:  N2 (g) + 3H2 (g) <---> 2NH3 (g)  Both reactants and products are gases, therefore it affect Kp value. This equation reaches equilibrium as it is in a "sealed container", also known as closed system. steps to find Kp:  1) Find the mole of reactants and product at equilibrium: From the stoichiometry equation, we can tell that 3 mol of Hydrogen is needed for every 1 mol of nitrogen. This means that for every 0.450 mol of N2, 1.35 mol of H2 is needed. Since we only have 0.450 mol of hydrogen, we can tell that hydrogen is the limiting reagent. 2) Draw a table to find equilibrium moles.                      d                      N      3H2      2NH3 Initial mole/ mol        0.450    0.450     0.000 Reacting mole/ mol     -0.050   -0.150    +0.100 Equilibrium mole/ mol    0.400     0.300     0.100 3) Calculate partial pressure for each of these 3 gases: N2= (0.4/ 0.8) X 500 kPa= 250 kPa H2= (0.3/0.8) X 500 kPa= 187.5 kPa NH3= (0.1/0.8) X 500kPa= 62.5 kPa 4) place this partial pressure into Kp equation: Kp = (PNH) / (PH2 ) (PN2 ) Kp= 2.37 X 10-6 kPa 

Answered by Nayan G. Chemistry tutor

11474 Views

See similar Chemistry A Level tutors

Related Chemistry A Level answers

All answers ▸

The ionic product of water, Kw = 2.93 × 10−15 mol dm−6 at 10 °C. Calculate the pH of a 0.0131 mol dm−3 solution of calcium hydroxide at 10 °C Give your answer to two decimal places.


1. Why does the first ionisation energy of atoms generally increase across a period?


How do I calculate the Gibbs free energy change for a reaction when given the entropy change, enthalpy change and the reaction conditions>


Which compound has a higher boiling point and why: water (H2O) or methane (CH4)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy