Question7 For the chemical reaction the equilibrium constant expression is NOBr

Question

Question7 For the chemical reaction the equilibrium constant expression is NOBr For this reaction, the value of K at 100 °C is 0.014 Initially 1.00 mol of NOBr is present and no Br or NO. From mass balance and stoichiometry, the concentrations at equilibrium are [NOB 1.00-2x, [Br, INO] 2 mol L. When substituted into the expression for K we get 1-4x+4 (a) Estimate [NO] at 100 °C using Newton's method. Use the plot (where ix)-4x- 0.056 +0.056x-0.014) and the table below to choose your starting x values. Calculate TWO iterations only. fx) 32.35 1.5 13.72 4.126 -0.5 -0.556 0.014 40 30 2D 10 -3 10 -20 30 3.986 13,444 231.874 (b) Explain how you would determine a more accurate value of [NOl using Newton's

Explanation / Answer

The equation is f(x)= 4x3-0.056x2+0.056x-0.014 (1)

Newton's method is x(i+i)= xi- f(xi)/f'(xi) (2)

f'(x)= df/dx =12x2-2*0.056x+0.056 =12x2-0.112x+0.056 (3)

xi= initial guess value and (i+i)= subsequenct value calculated from Eq.2

let the initial guess =0, f(x)=-0.014 and f'(x)= 12*0-0.056*0+0.056

x(i+1)= 0 -(-0.014/0.056)= 1/4=0.25

this will be the new value of x and this procedure will have to continue till the difference between two successive interations will be close to zero.

this will be new x value. x(0.25)= 0.0756, f'(x)= 0.778

( from Eq.2) ,x(i+1)= 0.25-0.0756/0.778 =0.196

f(0.196)= 0.03807, f'(0.1516)=0.495, x(i+1)= 0.196-0.0169/0.3142 =0.12

f(0.12)=0.006, f'(0.12)=0.2154, x(I+1)= 0.12-0.006/0.2154=0.092

f(0.092)=-0.00058, f'(0.092)=0.147, x(I+1)= 0.092+0.00058/0.147=0.0959

f(0.0959)= 0.000269, f'(0.0959)= 0.1556, x(i+1)= 0.0959-0.000269/0.1556 = 0.0942

f(0.0942)=-0.00011, f'(0.0942)= 0.1591, x(i+1)= 0.0942+0.00011/0.1591=0.0948

f(0.0948)= 2.55*10-5, f'(0.0948)=0.1532, x*(i+1)= 0.0948-2.55*10-5/ 0.0948= 0.0946

the difference between 0.0946 and 0.0948 is0.0002 which is very less .Hence the value os 0.0946

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