x=K=4.3x10 1.0-x x2=4.3×10-1.0-x) x2+4.3×10-7x-4.3×10-7=0 a=1b=4.3×10-7:c=-4.3×10-7 x=-b±v6-4ac 2a -4.3x10'±V85x109-40-43x10 21) -4.3×107±1.31×10-3 =-6.55×104or6.55×10 Thus,x=[HCO3(ao]=[H3O'(g]=6.55x 10mol L pH=-log10H,0aml=-log10(6.55x10)=3.18 HCO3(+H2OdIICO3+H3O K.-CO.H [HCO,(m】 or.CO5.6x10molL [H,0】 K 1.0×10-4 [oH✉小FH,0】6.55x10 =1.53×10-1"molL 3.For the reaction 2 NON+)the value of Kp at 1000C is 1.78x 10.If 1.00 atm of NO is placed in a flask at 1000C.calculate the equilibrium partial pressures(atm)of all three gases p(NO(g),atm p(N2(g)),atm p(O2(g)),atm Initial 1.00 0 0 Change -2X +x 5 1 7 a 2 7 27 7 7 7 2 7 13 7 3 4 4 x(x) K 4.3 10 1.0 x x 4.3 10 (1.0 x) x 4.3 10 x 4.3 10 0 a 1; b 4.3 10 ; c 4.3 10 b b 4ac x 2a 4.3 10 1.85 10 4(1)( 4.3 10 ) 2(1) 4.3 10 1.31 10 2 6.55 10 or 6.55 10 − − − − − − − − − − − − = =× − =× − +× −× = = = × =− × −± − = −× ± × − −× = −× ± × = =− × × −7 Thus, x = [HCO3 - (aq)] = [H3O+ (aq)] = 6.55 x 10-4 mol L-1 pH = -log10[H3O+ (aq)] = -log10(6.55 x 10-4) = 3.18 HCO3 - (aq) + H2O(l) ∏ CO3 -2 (aq) + H3O+ (aq) 2 2 2 2 3 (aq) 3 (aq) a 3 (aq) 2 1 a 3 (aq) 3 (aq) a 3 (aq) [CO ][H O ] K [HCO ] K [HCO ] or, [CO ] K 5.6 10 mol L [H O ] − + − − − − + = = == × 1 −1 14 w 11 1 (aq) 4 3 (aq) K 1.0 10 [OH ] 1.53 10 mol L [H O ] 6.55 10 − − − + − × = = =× × − 3. For the reaction 2 NO(g) ∏ N2(g) + O2(g), the value of Kp at 1000o C is 1.78 x 107 . If 1.00 atm of NO(g) is placed in a flask at 1000o C, calculate the equilibrium partial pressures (atm) of all three gases. p(NO(g)), atm p(N2(g)), atm p(O2(g)), atm Initial 1.00 0 0 Change -2x +x +x 5