01:18 Dim. 2 avr. 64 % moodle.concordia.ca a The standard Gibbs energy of formation of PH3(g) is +13.4 kJ mol-1 at 298 K. What is the corresponding reaction Gibbs energy when the partial pressures of the H2 and PH3 (treated as perfect gases) are 1.0 bar and 0.60 bar, respectively? Give your answer in kJ/mol Answer: 13.4 X Check Incorrect Marks for this submission: 0.00/2.00. ThisThe equilibrium constant for the reaction N2(g) + 02(g) = 2 N0(g) is 1.69 X 10'3 at 2300 K. A mixture consisting of 5.0 g of nitrogen and 2.0 g of oxygen in a container of volume 1.0 dm3 is heated to 2300 K and allowed to come to equilibrium. Calculate the mole fraction of N0 at equilibrium. Give your answer in kJ/mol Answer: 0.0403 X Check Solving the quadratic equation & ignoring the negative value, Data Page No. 30 = 0-0021216 [NOT = ZX = 2x 0-0021216 = 0 0042432 mal Love [ N ] = 5 - = 5-0-0021216 = 0-17644983 mal 28 28 e [Oz Jeq: = 1 -X = 1-0.0021216= 0-0603784 mol 16 16 Hence n No, 29 [ NO]. x1 2 = 0.0042432 model holes [ No ] x 18 = 0.17644983 moles = [02] x 12 = 0.0603784 moled " Mole fraction of NO at egi mof No q 0. 0042432 0 .0042432 + 0.176 44983 + 0 0603784 0.0042432 0.24107143 0 . 01760 14 22 (Aws )N + 02 = 2 NO Date * = 169 x 103 ( at 2300 k ) Page No. Intal = 59 mo = 2g Mall AA = 289 -, Moz = 32g 2 moles Initial 28 32 16 Moles Now Vol. of container = 1 dm' = Le .. [N, ] = /0 -= 5 molle 28 Initial lly [ 0, ] = I molle 16 N + 02 - 2NO Initial Cone 5/ 28 + 2x Equilibrium NAMING Cone 5- 2 x 28 16 We have K = 1. 69 x 10 ( 2 x ) 2 (5 - x ) ( 1 - 20) 162 Pug + 3 H2 () = 2 PM ,) Date A G = + 13 . 4 KJ / mal Page No. T - 298 K PH. = 1 bar PPM, 0:6 both Ah = AG + RT InQ = A ho + RT en ( Ppmy )2 13.4+ 8.314 x 298 In (0.6) 1000 = 13.4 + 0 2477.572 & (0.36) 1000 10. 868785 KJ/ mol ( And. )