In a given experiment 1.00 mole of N2 and 3.00 moles of H2 were placed in a 3.00

Question

In a given experiment 1.00 mole of N2 and 3.00 moles of H2 were placed in a 3.00L container. Equilibrium was established given the following N2(g) + 3H2(g) 2NH3(g) Complete the following table. Use numerical values in the Initial row and values containing the variable "x" in both the Change and Equilibrium reaction rows. Let x the amount of N2 needed to reach equilibrium. Do not use spaces in expressions containing a + or-sign. 2 2 NH3 Initial 0.333 Change Equilibrium 0.333-x At equilibrium the concentration of NH is 0.0521 M Determine the following quantities at equilibrium. Calculate the concentrations to 2 significant figures IN2l [H2] = Determine the numerical value of K for this reaction. Enter your answer to two significant figures.) Submit Show Hints

Explanation / Answer

initially

N2 = 0.333

H2 = 3/3 = 1 M

NH3 = 0

the change

N2 = -x

H2 = -3x

NH3 =+2x

in equilbirium

N2 = 0.333 - x

H2 = 1-3x

NH3 = 2x

if

[NH3] = 0.0521 --> x = 0.0521 /2 ;

x = 0.02605

N2 = 0.333 - 0.02605 = 0.30695 M

H2 = 1-3*0.02605 = 0.92185 M

now..

K = [NH3]^2 /([N2][H2]^3)

K = (0.30695 ^2) / ((0.30695 )*(0.92185 ^3))

K = 0.3918

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