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Get Full Access to Chemical Principles - 8 Edition - Chapter 10 - Problem 10.82
Get Full Access to Chemical Principles - 8 Edition - Chapter 10 - Problem 10.82

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# Consider the following reaction at 800. K: N2(g) 1 3F2(g) ISBN: 9781305581982 176

## Solution for problem 10.82 Chapter 10

Chemical Principles | 8th Edition

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Problem 10.82

Consider the following reaction at 800. K: N2(g) 1 3F2(g) 88n 2NF3(g) An equilibrium mixture contains the following partial pressures: PN2 5 0.021 atm, PF2 5 0.063 atm, and PNF3 5 0.48 atm. Calculate DG8 for the reaction at 800. K.

Step-by-Step Solution:

March 21­25, 2016 Section 3.2 Suppose f’(x) >0 for all x on an open interval I. Suppose x < x i1 I. 2 en f(x) is continuous on [x1 x2] and it is differentiable on (x 1 x 2 2 x ¿ So by Mean Value Theorem there is c ¿ f( )1f ¿ ' ϵ ( 1x S2)hthen f (c)=¿ Which implies Thus f(x)< 0 is increasing on I. By similar arguments, we can show that f’(x)< 0 on an open interval f is decreasing on I. And if f’(x)=0 on an open interval I, then f(x) is constant on I. Theorem If f’(x)= g’(x) on an open interval I, then g(x)= f(x) + c for some constant c. 3 2 Sketch curve y= x ­3x + 7x­2 Step 1 Find Critical points ' 2 y =3x −6 x+7 2 (−6) −4 3 )7 )<0 No real roots y’ >0 for all real x So y is increasing 3 2 Find the root for y= x ­3x + 7x­2 Can we estimate the root Since y(0)= ­2 y(1)=3 The root is between 0 and I, not a rational root so we use technology to approximate. ‘ ‘ ‘ ‘ ‘ ‘ ‘ ¿3x−2 Domainis −∞,∞ ) y x +1 2 3 (x +1 − (x−2 2 )( ) y’(x)= 2 2 x +1 ) −3x +4x+3 −(3x −4x−3) 2 = 2 = x +1 ) (x +1) 2 Let 3x −4x−3=0 2 ´ 4± √ −4 3 ( )( ) 2± √3 x= 2(3) = 3

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##### ISBN: 9781305581982

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