Question #8697a

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Answer 1 · 2016-04-29T16:19:53

$K_c=6.55xx10^(14)R$

$2NO_((g))+O_(2(g))rightleftharpoons2NO_(2(g))$

$K_p=(p_(NO_2)^(2))/(p_(NO)^(2)xxp_(O_2))$

$p$ represents the partial pressure of the gas.

The relationship between $K_p$ and $K_c$ is given by:

$K_p=K_c(RT)^(Deltan)$

$R$ is the gas constant

$T$ is the absolute temperature

$Deltan$ is the difference between the no. moles of product molecules - the number of moles of reactant molecules.

In this case you can see that $Deltan=2-3=-1$

$:.K_p=K_c(RT)^(-1)$

$K_p=K_c/(RT)$

$:.K_c=K_pxxRT$

$K_c=2.2xx10^(12)xxRxx298$

$K_c=6.55xx10^(14)R$

You can't evaluate this numerically since no units are given for $K_p$ so you can't chose a value for $R$ .