3.16可逆反应 →>B B 力界条件t=0cA=CCB=CBo t=t XC=Cpo +X d(cAo k,(Co-x)+k(cBo+x dt dx k (Cox)-k(cBo+x =(-k1x-k1x)+(k1 A0 BO (h,x+k-x)+(k,co-k3.1.6 可逆反应 边界条件:t=0 t=t A B ⎯⎯k1→ A B ⎯k−1 ⎯ A A0 c = c B B0 c = c c c x A = A0 − c c x B = B0 +   = − + + − = − − + − = − − + = − − + + − − − − − − − x t A B A B A B A B A dt k x k x k c k c dx k x k x k c k c k c x k c x dt dx k c x k c x dt d c x 0 0 1 1 1 0 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 0 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )