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2 October, 19:11

Consider the reaction: 2no (g) + br2 (g) ⇌2nobr (g) kp=28.4 at 298 k in a reaction mixture at equilibrium, the partial pressure of no is 102 torr and that of br2 is 160 torr. part a what is the partial pressure of nobr in this mixture? express your answer using three significant figures.

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  1. 2 October, 22:56
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    From the Equilibrium constant we have K_p = (P_Nobr) ^2 / (P_No) ^2 * (P_Br)

    K_p = 28.4. P_Br = 160. P_Nobr = x P_No = 102

    So we have 28.4 = x^2 / 160 * 102

    x = âš 28.4 * 160 * (102) ^2 = âš47275776 = 6875.738

    To 3 Sf becomes 6880.000
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