WWN9545: Work 2 X C Chemistry question I Che... X ii 3 smartwork. wwnorton.com /

Question

WWN9545: Work 2 X C Chemistry question I Che... X ii 3 smartwork. wwnorton.com /sw/mod/smartwork/view.php?idJ515836&isStudent; Search SMARTWORK 3.0 WWN9545 SMARTWORK #2 (DUE JAN. 31, 59 PM lil Ebook dic Table Question 4 of 8 (1 point) The value of Kp for the reaction 3 Hale) N2(g) A 2 NH g) is 4.3 x10 4 at 648 K. Determine the equilibrium partial pressure of NH3 in a reaction vessel that initially contained 0.900 atm N2 and 0.500 atm H2 at 648 K. Number MATH HINT: The equilibrium constant is very small atm v Check Answer Q View Solution Hint 4 5 6 7 8 2 3 Progress Exit Activity Activity Grade: 8 Points Possible

Explanation / Answer

Answer – We are given, reaction –

3H2(g) + N2(g) <----> 2 NH3(g)

Kp = 4.3*10-4 , T = 648 K , P of N2(g) = 0.900 atm , P of O2 = 0.500 atm

First we need to put the ICE table

We need to put ICE table

       3H2 (g) + N2 (g) <-----> 2 NH3(g)

I    0.500        0.900                  0

C    -3x           -x                     +2x

E 0.500-3x   0.900-x              +2x

Kp = P of (NH3(g))2 / P of (H2 (g))3 * P of (N2 (g))

4.3*10-4 = (2x)2 /(0.500-3x)3 *(0.900-x)

0.01161x4-0.0162x3+0.00619x2-0.000924+0.0000484 = 4x2

0.01161x4-0.0162x3- 3.99 x2-0.000924+0.0000484 = 0

By solving the quadratic equation

x = 0.00374

so at the equilibrium

partial pressure of NH3(g) = 2x

                                          = 2* 0.00374

                                          = 0.00748 atm

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