Question #353c0

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Answer 1 · 2017-03-17T05:27:34

$K_p = 2.25$

$Delta_r G= Delta_r G^0 + RT ln Q$

The reaction is occurring at standard conditions then $Delta_r G^0=0$ and $Q= K_(eq)$

$Delta_r G= RT ln Q$

$176.31 = 0.0821 xx 298 xx ln Q$

$ln Q = 7.2063$

$Q =1347.895$

$Q =K_(eq)=1347.895$

$K_p = K_c (RT)^(Delta ng)$

$K_p = 1347.895xx (0.0821 xx 298)^(-2)$

$K_p = 1347.895xx (24.4658)^(-2)$

$K_p = 1347.895/ (24.4658)^(2)$

$K_p = 1347.895/ 598.5753$

$K_p = 2.2518$

$K_p = 2.25$