path: root/Physical_And_Chemical_Equilibrium_For_Chemical_Engineers/ch12.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 12 : Chemical Equilibrium"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.1   Page: 311\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "T = 298.15      #[K] temperature\n",
      "P = 1.          #[atm] pressure \n",
      "R = 8.314*10**(-3)      #[kJ/(mol*K)]\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "g_a_0 = -20.9       #[kJ/mol]\n",
      "g_b_0 = -17.2       #[kJ/mol]\n",
      "\n",
      "x_a = math.exp((g_b_0-g_a_0)/(R*T))/(1+math.exp((g_b_0-g_a_0)/(R*T)))\n",
      "x_b = 1-x_a\n",
      "\n",
      "print \" The chemical equilibrium composition of the gaseous mixture contains %f mol fraction isobutane \\t\\t\\t\\t\\t\\t\\t\\tan %f mol fraction n-bumath.tane\"%(x_a,x_b)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The chemical equilibrium composition of the gaseous mixture contains 0.816475 mol fraction isobutane \t\t\t\t\t\t\t\tan 0.183525 mol fraction n-bumath.tane\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.2   Page: 319\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "T = 298.15              #[K] temperature\n",
      "P = 0.987               #[atm] pressure\n",
      "g_0_NO = 86.6           #[kJ/mol] Free energy of formation the NO from elements\n",
      "R = 8.314               #[J/(mol*K)]\n",
      "\n",
      "g_0_O2 = 0.00\n",
      "g_0_N2 = 0.00\n",
      "\n",
      "\n",
      "delta_g_0 = 2*g_0_NO - g_0_O2 - g_0_N2          #[kJ/mol]\n",
      "delta_g_01 = delta_g_0*1000                     #[J/mol]\n",
      "\n",
      "K_298 = math.exp((-delta_g_01)/(R*T))\n",
      "\n",
      "f_0_N2 = 1.             #[bar]\n",
      "f_0_O2 = 1.             #[bar]\n",
      "f_0_NO = 1.             #[bar]\n",
      "\n",
      "\n",
      "\n",
      "y_N2 = 0.78\n",
      "y_O2 = 0.21\n",
      "\n",
      "y_NO_298 = math.sqrt(K_298*y_N2*y_O2)\n",
      "\n",
      "T_1 = 2000                  #[K]\n",
      "K_2000 = 4.0*10**-4\n",
      "\n",
      "y_NO_2000 = math.sqrt(K_2000*y_N2*y_O2)*10**(6)         #[ppm]\n",
      "\n",
      "print \" The equilibrium constant for the reaction at 298.15 K is \\t\\t\\t %.1e\"%(K_298)\n",
      "print \" The concentration of NO at equilibrium at temperature 298.15 K is \\t\\t %.1e\"%(y_NO_298)\n",
      "print \" The equilibrium consmath.tant for the reaction at 2000 K is \\t\\t\\t %.1e\"%(K_2000)\n",
      "print \" The concentration of NO at equilibrium at temperature 2000 K is \\t\\t %.0f ppm\"%(round(y_NO_2000,-2))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The equilibrium constant for the reaction at 298.15 K is \t\t\t 4.5e-31\n",
        " The concentration of NO at equilibrium at temperature 298.15 K is \t\t 2.7e-16\n",
        " The equilibrium consmath.tant for the reaction at 2000 K is \t\t\t 4.0e-04\n",
        " The concentration of NO at equilibrium at temperature 2000 K is \t\t 8100 ppm\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.3   Page: 321\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "Temp = 2000.                #[K]\n",
      "n_air = 1.                  #[mol] no of moles of the air\n",
      "\n",
      "\n",
      "K_2000 = 4*10**(-4)\n",
      "\n",
      "def f(x): \n",
      "\t return  (2*x)**(2) - K_2000*(0.78-x)*(0.21-x)\n",
      "\t #return  (K_2000-2)*x**(2)-K_2000*(0.78+0.21)*x+K_2000*0.78*0.21\n",
      "x = fsolve(f,0)\n",
      "c_NO = 2*x*10**(6)              #[ppm]\n",
      "p = c_NO/8100.*100\n",
      "\n",
      "print \" The calculated NO cocentration is %f ppm, which %f%% of the value computed in example 12.1\"%(c_NO,p)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The calculated NO cocentration is 7996.442873 ppm, which 98.721517% of the value computed in example 12.1\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.5   Page: 324\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "Temp = 298.             #[K]\n",
      "K = 29.6                # equilibrium consmath.tant at 298 K \n",
      "P = 1.                  #[bar]\n",
      "n_water_0 = 0.833       #[mol]\n",
      "n_ethylene_0 = 1.       #[mol]\n",
      "n_ethanol_0 = 0.        #[mol]\n",
      "\n",
      "n_T_0 = (n_water_0+n_ethylene_0+n_ethanol_0)        #[mol]\n",
      "\n",
      "def f(e): \n",
      "\t return  ((0+e)/(1.833-e))/(((1-e)/(1.833-e))*((0.833-e)/(1.833-e)))-K*P/(1)\n",
      "e_1 = fsolve(f,0)\n",
      "e_2 = fsolve(f,0.5)\n",
      "\n",
      "y_ethanol = ((0+e_2)/(1.833-e_2))\n",
      "y_ethylene = ((1-e_2)/(1.833-e_2))\n",
      "y_water = ((0.833-e_2)/(1.833-e_2))\n",
      "\n",
      "print \"Concentration of the ethylene at the equilibrium is %f\"%(y_ethylene)\n",
      "print \" Concentration of the water at the equilibrium is    %f\"%(y_water)\n",
      "print \" Concentration of the ethanol at the equilibrium is  %f\"%(y_ethanol)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Concentration of the ethylene at the equilibrium is 0.243702\n",
        " Concentration of the water at the equilibrium is    0.092079\n",
        " Concentration of the ethanol at the equilibrium is  0.664219\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.6   Page: 324\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "Temp = 273.15+25            #[C]\n",
      "P = 1.                      #[bar]\n",
      "R = 8.314                   #[J/(mol*K)]\n",
      "\n",
      "g_H2O_0 = -237.1            #[kJ/mol]\n",
      "g_O2_0 = 0                  #[kJ/mol]\n",
      "g_H2_0 = 0                  #[kJ/mol]\n",
      "\n",
      "delta_g_0 = g_H2O_0 - 0.5*g_O2_0-g_H2_0         #[kJ/mol]\n",
      "delta_g_1 = delta_g_0*1000                      #[J/mol]\n",
      "\n",
      "K = math.exp((-delta_g_1)/(R*Temp))\n",
      "\n",
      "n_T_0 = 1.5#[mol]\n",
      "\n",
      "\n",
      "def f(e): \n",
      "\t return (e/(1.5-0.5*e))/(((1-e)/(1.5-0.5*e))*((0.5-0.5*e)/(1.5-0.5*e))**(0.5))\n",
      "\n",
      "y_H2 = 2.4e-28\n",
      "y_O2 = 0.5*y_H2\n",
      "\n",
      "print \" The equilibrium mol fraction of the hydrogen is   %0.3e\"%(y_H2)\n",
      "print \" And the equilibrium mol fraction of the oxygen is %e \"%(y_O2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The equilibrium mol fraction of the hydrogen is   2.400e-28\n",
        " And the equilibrium mol fraction of the oxygen is 1.200000e-28 \n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.7   Page: 327\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "Temp = 298.15           #[K]\n",
      "Press = 1*10**(5)       #[Pa]\n",
      "R = 8.314               #[J/(mol*K)]\n",
      "\n",
      "v_liquid = 1.805*10**(-5)           #[m**(3)/mol] this liquid specific volume and we will treat is as a consmath.tant\n",
      "\n",
      "P_vapour_25 = 0.0317*10**(5)            #[Pa]\n",
      "\n",
      "\n",
      "def f0(P): \n",
      "\t return v_liquid*P**(0)\n",
      "\n",
      "delta_g_0_1 =  quad(f0,Press,P_vapour_25)[0]\n",
      "\n",
      "\n",
      "delta_g_0_2 = 0             #[J/mol]\n",
      "\n",
      "\n",
      "def f1(P): \n",
      "\t return 1./P\n",
      "\n",
      "delta_g_0_3 = (R*Temp)* quad(f1,P_vapour_25,Press)[0]\n",
      "\n",
      "\n",
      "delta_g_0 = delta_g_0_1+delta_g_0_2+delta_g_0_3         #[J/mol]\n",
      "delta_g_1 = delta_g_0/1000                              #[kJ/mol]\n",
      "\n",
      "print \" Total change in the free energy of water going under given conditions is %0.2f kJ/mol\"%(delta_g_1)\n",
      "\n",
      "delta_g_0_ideal_gas = -228.6                            #[kJ/mol]\n",
      "delta_g_0_liquid = -237.1                               #[kJ/mol]\n",
      "delta_g_o = delta_g_0_ideal_gas-delta_g_0_liquid        #[kJ/mol]\n",
      "\n",
      "print \" From the values of Table A.8 given in the book the free energy change is %0.2f kJ/mol\"%(delta_g_o)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Total change in the free energy of water going under given conditions is 8.55 kJ/mol\n",
        " From the values of Table A.8 given in the book the free energy change is 8.50 kJ/mol\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.8   Page: 330\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from scipy.integrate import quad \n",
      "\n",
      "T1 = 273.15+25              #[K]\n",
      "T2 = 273.15+400             #[K]\n",
      "R = 8.314                   #[J/(mol*K)]\n",
      "\n",
      "g0_NH3 = -16.5              #[kJ/mol]\n",
      "g0_N2 = 0                   #[kJ/mol]\n",
      "g0_H2 = 0                   #[kJ/mol]\n",
      "\n",
      "\n",
      "delta_g_0 = g0_NH3 - 0.5*g0_N2 - 1.5*g0_H2          #[kJ/mol]\n",
      "\n",
      "K_1 = math.exp(-delta_g_0*1000/(R*T1))          # Equilibrium consmath.tant of the reaction at temperature 298.15 K\n",
      "\n",
      "h0_NH3 = -46.1          #[kJ/mol]\n",
      "h0_N2 = 0               #[kJ/mol]\n",
      "h0_H2 = 0               #[kJ/mol]\n",
      "\n",
      "del_h_1 = h0_NH3 - 0.5*h0_N2 - 1.5*h0_H2        #[kJ/mol]\n",
      "\n",
      "a_NH3 = 3.578\n",
      "a_H2 = 3.249\n",
      "a_N2 = 3.280\n",
      "b_NH3 = 3.020*10**(-3)      #[1/K]\n",
      "b_H2 = 0.422*10**(-3)\n",
      "b_N2 = 0.593*10**(-3)\n",
      "c_NH3 = 0                   #[1/K**(2)]\n",
      "c_H2 = 0                    #[1/K**(2)]\n",
      "c_N2 = 0                    #[1/K**(2)]\n",
      "d_NH3 = -0.186*10**(5)      #[K**(2)]\n",
      "d_H2 = 0.083*10**(5)        #[K**(2)]\n",
      "d_N2 = 0.040*10**(5)        #[K**(2)]\n",
      "\n",
      "del_a = a_NH3 - 0.5*a_N2 - 1.5*a_H2\n",
      "del_b = b_NH3 - 0.5*b_N2 - 1.5*b_H2\n",
      "del_c = c_NH3 - 0.5*c_N2 - 1.5*c_H2\n",
      "del_d = d_NH3 - 0.5*d_N2 - 1.5*d_H2\n",
      "\n",
      "I = R*( del_a*T1 + del_b*T1**(2)/2 + del_c*T1**(3)/3 - del_d/T1)        #[J/mol]\n",
      "\n",
      "\n",
      "def f5(T): \n",
      "\t return (del_h_1*1000 - I + R*(del_a*T + del_b*T**(2)/2 + del_c*T**(3)/3 - del_d/T))/T**(2)\n",
      "\n",
      "X = (1/R)* quad(f5,T1,T2)[0]\n",
      "\n",
      "\n",
      "K_2 = K_1*math.exp(X)\n",
      "\n",
      "print \" Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are \"\n",
      "print \" K = %0.0f at temperature 25 deg C\"%(K_1)\n",
      "print \" K = %f at temperature 400 deg C\"%(K_2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Equilibrium consmath.tants for the formation of ammonia from hydrogen and nitrogen are \n",
        " K = 778 at temperature 25 deg C\n",
        " K = 0.013588 at temperature 400 deg C\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.9   Page: 335\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "n_H2_0 = 1.5            #[mol]\n",
      "n_N2_0 = 0.5            #[mol]\n",
      "n_NH3_0 = 0             #[mol]\n",
      "T_1 = 298.15            #[K]\n",
      "T_2 = 673.15            #[K]\n",
      "P = 1.                  #[bar]\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "K_298 = 778.            # at temperature 298.15K\n",
      "K_673 = 0.013           # at temperature 673.15K\n",
      "\n",
      "def g(e_1): \n",
      "\t return  ((0+e_1)/(2-e_1))/(((0.5-0.5*e_1)/(2-e_1))**(0.5)*((1.5-1.5*e_1)/(2-e_1))**(1.5))-K_298\n",
      "     \n",
      "e_1 = fsolve(g,0.97)\n",
      "y_NH3_298 = e_1/(2-e_1)\n",
      "\n",
      "def h(e_2): \n",
      "\t return  ((0+e_2)/(2-e_2))/(((0.5-0.5*e_2)/(2-e_2))**(0.5)*((1.5-1.5*e_2)/(2-e_2))**(1.5))-K_673\n",
      "\n",
      "e_2 = fsolve(h,0)\n",
      "y_NH3_673 = e_2/(2-e_2)\n",
      "\n",
      "print \" The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is %f\"%(y_NH3_298)\n",
      "print \" The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is %f\"%(y_NH3_673)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The mole fraction of NH3 in the equilibrium at the temperature 298.15K and 1 bar is 0.939036\n",
        " The mole fraction of NH3 in the equilibrium at the temperature 673.15K and 1 bar is 0.004187\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.10  Page: 337\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "Temp = 273.15+400           #[K]\n",
      "P = 150*1.01325             #[bar]\n",
      "\n",
      "K_673 = 0.013\n",
      "\n",
      "K = K_673*(P/1)**(1.5+0.5-1)\n",
      "\n",
      "def f(e): \n",
      "\t return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K\n",
      "e=fsolve(f,0.5)\n",
      "\n",
      "y_NH3 = (0+e)/(2-e)\n",
      "\n",
      "print \"The mole fraction of the ammonia in the equilibrium is %0.2f\"%(y_NH3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mole fraction of the ammonia in the equilibrium is 0.31\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.11   Page: 338\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "T = 273.15+400              #[K] given temperature\n",
      "P = 150*1.01325             #[bar] given pressure\n",
      "\n",
      "K_673 = 0.013\n",
      "\n",
      "K_v = (10**((0.1191849/T + 91.87212/T**(2) + 25122730/T**(4))*P))**(-1)\n",
      "\n",
      "K = (K_673/K_v)*(P/1)**(1.5+0.5-1)\n",
      "\n",
      "\n",
      "def f(e): \n",
      "\t return  ((0+e)/(2-e))/(((0.5-0.5*e)/(2-e))**(0.5)*((1.5-1.5*e)/(2-e))**(1.5))-K\n",
      "e = fsolve(f,0.2)\n",
      "\n",
      "y_NH3 = (0+e)/(2-e)\n",
      "\n",
      "print \" The mole fraction of the ammonia in the equilibrium is %0.2f\"%(y_NH3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The mole fraction of the ammonia in the equilibrium is 0.34\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.12   Page: 340\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "from scipy.optimize import fsolve \n",
      "import math \n",
      "\n",
      "p_i = 1.                #[atm] initial pressure \n",
      "P = 150.                #[atm] final pressure\n",
      "T = 273+25.             #[K] Given temperature\n",
      "R = 8.314               #[J/(mol*K)]\n",
      "\n",
      "delta_g_0 = 10.54*1000          #[J/mol]\n",
      "\n",
      "K = math.exp((-delta_g_0)/(R*T))\n",
      "\n",
      "\n",
      "\n",
      "def f(e): \n",
      "\t return  (e*e)/((1-e)*(1-e))-K\n",
      "e = fsolve(f,0)\n",
      "\n",
      "v_C2H5OOC2H5 = 97.67            #[ml/mol]\n",
      "v_H2O = 18.03                   #[ml/mol]\n",
      "v_C2H5OH = 58.30                #[ml/mol]\n",
      "v_CH3COOH = 57.20               #[ml/mol]\n",
      "\n",
      "delta_v = v_C2H5OOC2H5 + v_H2O - v_C2H5OH - v_CH3COOH       #[ml/mol]\n",
      "\n",
      "\n",
      "\n",
      "def g(e_1): \n",
      "\t return  (e_1*e_1)/((1-e_1)*(1-e_1))*math.exp((delta_v)/(R*T)*(P-p_i))-K\n",
      "e_1 = fsolve(g,0.2)\n",
      "\n",
      "a = e_1/e\n",
      "\n",
      "\n",
      "print \"On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar \\\n",
      "reacmath.tants at equilibrium becomes %f times of initial\"%(a)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "On increamath.sing the pressure from 1 atm to 150 atm the reacted amount of the equimolar reacmath.tants at equilibrium becomes 0.994639 times of initial\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      " Example 12.13  Page: 342\n"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "P = 150.            #[atm] given pressure\n",
      "T = 400.            #[C] temperature\n",
      "K = 0.013\n",
      "K_v = 0.84\n",
      "delta_v = 1.5+0.5-1\n",
      "\n",
      "\n",
      "K_p = (K/K_v)*(1/1)**(delta_v)              #[1/bar]\n",
      "\n",
      "print \" Value of the K_p at the given condition is %f 1/bar)\"%(K_p)\n",
      "print  \" The basic K is dimensionless%( but K_p has the dimensions of pressure to the power.\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Value of the K_p at the given condition is 0.015476 1/bar)\n",
        " The basic K is dimensionless%( but K_p has the dimensions of pressure to the power.\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}