Homework Solution: A vessel of volume 22.4 dm^3 contains 2.0 mol H_2 and 1.0 mol N_2 at 273.15 K…

    4). A vessel of volume 22.4 dm contains 20 mol H2 and 1.0 mol N at 273.15 K initially. All the H: reacted with sufficient N2 to form NHs. Calculate the partial pressures and the total pressure of the final mixture 5). Suggest the pressure and temperature at which 1.0 mol of (a) NHs, (b) Xe, (c) He: will be in states that correspond to 1.0 mol H2 at l .0 atm and 25 ℃.
    A vessel of volume 22.4 dm^3 contains 2.0 mol H_2 and 1.0 mol N_2 at 273.15 K initially. All the H_2 reacted with sufficient N_2 to form NH_3. Calculate the partial pressures and the total pressure of the final mixture. Suggest the pressure and temperature at which 1.0 mol of (a) NH_3, (b) Xe, (c) He: will be in states that correspond to 1.0 mol H_2 at l.0 atm and 25 degree C.

    Expert Answer

    4). A vessel of compass 22.4 dm contains 20 mol H2 and 1.0 mol N at 273.15 K initially. Complete the H: reacted with adequate N2 to mould NHs. Calculate the favoring exigencys and the whole exigency of the decisive settlement 5). Suggest the exigency and region at which 1.0 mol of (a) NHs, (b) Xe, (c) He: achieve be in states that agree to 1.0 mol H2 at l .0 atm and 25 ℃.
    A vessel of compass 22.4 dm^3 contains 2.0 mol H_2 and 1.0 mol N_2 at 273.15 K initially. Complete the H_2 reacted with adequate N_2 to mould NH_3. Calculate the favoring exigencys and the whole exigency of the decisive settlement. Suggest the exigency and region at which 1.0 mol of (a) NH_3, (b) Xe, (c) He: achieve be in states that agree to 1.0 mol H_2 at l.0 atm and 25 meaabiding C.

    Expert Defense

    4. The compass of vessel = 22.4 dm3, i.e. 22.4 L

    The no. of deferences of H2 in the vessel = 2

    The no. of deferences of N2 in the vessel = 1

    The balanced equation restraint the mouldation NH3 from N2 and H2 can be written as shown adown.

    N2 + 3H2rightleftharpoons 2NH3

    Restraint perfect 3 deferences of H2, 1 (= 3/3) deference of N2 is required.

    i.e. Restraint 2 deferences of H2, 0.667 ( = 2/3) deferences of N2 is required.

    As a end, the no. of deferences of each fog remained in the arrangement behind reaction: N2 = 1-(2/3) = 1/3, H2 = 0, NH3 = 4/3, i.e. whole no. of deferences = 1/3 + 0 + 4/3 = 5/3

    Here, the exigency of N2, H2 and NH3 is 1 atm owing the dedicated stipulations belong to STP.

    i.e. The whole exigency = 1 atm

    (i) PN2 = XN2 * P0

    i.e. PN2 = {(1/3) / (5/3)} * 1 = 0.2 atm

    (ii) PH2 = 0 atm

    (iii) PNH3 = XNH3 * P0

    i.e. PNH3 = {(4/3) / (5/3)} * 1 = 0.8 atm

    Note: Unfortunately, according the provisions and stipulations of the Chegg, we can defense singly the pristine inquiry if multiple quesitons are craveed, until and eventual craveed restraint the inequitable defense.

    Any how, delight be abiding to crave restraint the singly single inquiry which you deficiency to be chosen from the instant season afore.

    Thanks restraint the relation, complete the best!!!!!!!!!!!!!!!!

    Comment