Qn no 1 Qn no 1 R". 2017 1 MEDICAL as are measured at equilibrium N, • 038 atm. • atm Hydrvy-en is renuwed from the vessel until the partial pressure equal to 0 45 atm. calculate the partial pressures of the other of A. B and C for the reaction A B + C are 4.6. 2.3 and at 25'C. tf 2 rtx"es.'lt of A are removed calculated the equilibrium A S C at sarne temperature. A 2 vessel cmtains 0.48 mole of C02, 0.48 mole of H2, 0.96 mole of H20 and 0.96 moles Of CO at eou•rum C02(g) + H2(g) H20(g) + CO(g). (A) How many moles of H2 must be added to he concentraton of CO to 0.6 M. (B) How many moles of C02 must be added bring the CO cmæntrat07 to 0.6 M. (C) How many moles of H20 must be removed to bring the CO conæntraton 0.6 n roles each of H20, H2 and 02 are mixed at a suitable high temperature to attain the equilibrium + 02. If y moles of H20 are dissociated and the total pressure maintained as F Calculate Kp. A compound AB2(g) dissociates according to the reaction 2AB(g) + B2(g) Wth a degn of dissodaåon 'x'. vhich is small compared to unity. Deduce the expression for 'x' in terms of t equilibrium constant Kp and the total pressure P. CalctJlate the percent dissociation of H2S(g) if 0.1 mole of HS is kept in 0.4 It. vessel at 10000K. the reaction 2H2S(g) 2H2(g) + S2(g) the value of Kc is Share with your friends Share 1 Ashish Khanna answered this Dear Student The equilibrium reaction is as follows: CO2 + H2 ⇌ H2O + CO No. of moles 0.48 0.48 0.96 0.96at eqm. Conc (in 2L) 0.24M 0.24M 0.48M 0.48MEquilibrium constant k= COH2OCO2H2 =0.48 × 0.480.24× 0.24 = 4 Now, (A) Let the concentration of H2 be x So, k= COH2OCO2H2H2 = COH2Ok ×CO2 = 0.6 ×0.484 × 0.24 = 0.3 M So, the no. of moles of H2 will be 0.3 × 2 = 0.6 mole Hence to bring the concentration of CO to 0.6 M, the added number of moles of H2 will be = 0.6 - 0.48 = 0.12 mol (B) Let the concentration of CO2 be x So, k= COH2OCO2H2CO2= COH2Ok ×H2 = 0.6 ×0.484 × 0.24 = 0.3 M So, the no. of moles of CO2 will be 0.3 × 2 = 0.6 mole Hence to bring the concentration of CO to 0.6 M, the added number of moles of CO2 will be = 0.6 - 0.48 = 0.12 mol (C) Let the concentration of H2O be x So, k= COH2OCO2H2H2O= k ×H2CO2CO = 4 ×0.24×0.240.6 = 0.384 M So, the no. of moles of CO2 will be 0.384 × 2 = 0.768 mole Hence to bring the concentration of CO to 0.6 M, the removed number of moles of H2O will be = 0.96 - 0.768 = 0.192 mol Regards -1 View Full Answer
