Asked by Anna
The equilibrium constant kc for the reaction
N2(g)+ 3H(g)---2NH3(g)
at 450C is 0.159. Calculate the equilibrium composition when 1.00 mol N2 is mixed with 3.00 mol H2 in a 2.00-L vessel.
need help setting up the problem
N2(g)+ 3H(g)---2NH3(g)
at 450C is 0.159. Calculate the equilibrium composition when 1.00 mol N2 is mixed with 3.00 mol H2 in a 2.00-L vessel.
need help setting up the problem
Answers
Answered by
DrBob222
Correct the equation to
N2 + 3H2 ==> 2NH3
Set up an ICE chart.
initial:
N2 = 1 mole/2 L = 0.5 M
H2 = 3 moles/2 L = 1.5 M
NH3 = 0
change:
NH3 = +2x
N2 = -x
H2 = -3x
equilibrium:
N2 = 0.5-x
H2 = 1.5-3x
NH3 = 2x
Substitute into Kc expression and solve. Post your work if you get stuck.
N2 + 3H2 ==> 2NH3
Set up an ICE chart.
initial:
N2 = 1 mole/2 L = 0.5 M
H2 = 3 moles/2 L = 1.5 M
NH3 = 0
change:
NH3 = +2x
N2 = -x
H2 = -3x
equilibrium:
N2 = 0.5-x
H2 = 1.5-3x
NH3 = 2x
Substitute into Kc expression and solve. Post your work if you get stuck.
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