Given Kc = 1.5 × 1018 at 300 K for the reaction below 2 NO-----> N2 + ...

Given Kc = 1.5 × 10¹⁸ at 300 K for the reaction below

2 NO-----> N₂ + O₂

900 mg of NO were initially placed in a 1.00 L vessel and the reaction was allowed to reach equilibrium.Calculate the equilibrium concentrations of NO, N₂, and O₂.

Solutions

Expert Solution

Initial mass of NO = 900 mg = 0.9 g

Volume of vessel = 1.0 L

Molar mass of NO = 30.01 g/mol

Moles of NO = 0.9 / 30.01

= 0.0299 moles

Kc of reaction = 1.5 * 10¹⁸

                                      2 NO              N₂        + O₂
Initial                            0.0299                             0                         0
Change                         - 2x                               + x                      + x
Final                        0.0299 - 2x                           x                        x

Kc = [N₂][O₂] / [NO]²

1.5 * 10¹⁸ = x * x / (0.0299 - 2x)²

x = 0.01495

Hence, at equilibrium:

[NO] = 0.0299 - 2*0.01495

= 0 M

[N₂] = [O₂] = 0.01495M

queen_honey_blossom answered 1 year ago

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