Question

The equilibrium constant, K , for the following reaction is 1.80x104 at 298 K. NH_HS(s) NH3(g) + H :) Calculate the equilibrium concentration of H2S when 0.202 moles of NH HS(s) are introduced into a 1.00 L vessel at 298 K. [H2S] = M

Answer 1

& The equilibrium constante, ke for the following reaction is 1.80x10 at 28. 298k. NIGHS (S) & NH3(g) + H₂ SCO) 0.202 moles of NHGHS (s) are in into a Lool vessel at 298k. We know that, introduced NHOHS (s) are ke a concentration of product concentration of reaction reactant amowl [NH₃] [H₂S] [ NHg HS] 0.202 moles of NHaus is the initial amount of NHA HS. From this x'amount NH3 and axa of Hos are produced. i Remaining amount of Niats = ( 0.202-x) enke al 2 [Wlt3] [H2S] [Nha HS -4 xxx 0.7, 18X105 -4 22 or 0·202-2 1.8 x 10 = 12 0.202-2

& synthests o n a 097 3.636x10-5-8X10% = 22 or re²+ 1.8840 t1.8840 - 3.6 - 2.636 x 100 =0 on This is a quadratic equation (ax²+bx+c=o). The roots are - se a - b ± √b2 - 4ac where a=1 2a b=1.8x109 C2-36636x1534 --(2.8x10 ) + √(+8X209) ² 4 (1) (-3.636x15) 2 X 2-(8x10-3+ o. 012061193 2 - (1.8x109) +0.012061193 -(1.88 to 4 -0.0120.61193 2 0.011881193 -0.012241293 20.01188 (5.9405965x10), 1-6:1205965x103) Neglecting the negative value we get, n=5.940 6x 103

- The equilibrium concentration of Hos is 5.9406410 M.