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Question: For the equilibrium <img src="https://cdn.pureessence.tech/canvas_245.png?top_left_x=676&top_left_y=...

For the equilibrium ⇌2NH3,Kc2 \mathrm { NH } _ { 3 } , K _ { c } at 1000K is2.37×10−32.37 \times 10 ^ { - 3 }. If at equilibrium[N2]=2M,[H2]=3M\left[ N _ { 2 } \right] = 2 M , \left[ H _ { 2 } \right] = 3 M, the concentration of NH3\mathrm { NH } _ { 3 } is.

A

0.00358 M

B

0.0358 M

C

0.358 M

D

3.58 M

Answer

0.358 M

Explanation

Solution

2.37×10−3=x2[2][3]3=x2=0.127982.37 \times 10 ^ { - 3 } = \frac { x ^ { 2 } } { [ 2 ] [ 3 ] ^ { 3 } } = x ^ { 2 } = 0.12798

x = 0.358 M.