# Consider the equilibrium 2A <=>A2 Kp = 3.72 /atm at some temperature T K.The total equilibrium pressure = 1.5 atm.Calculate the partial pressure of the dimer

The computation is expressed below,

$2A\iff {A}_{2}\phantom{\rule{0ex}{0ex}}P0\phantom{\rule{0ex}{0ex}}P-pp\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}P=1.5atm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{K}_{p}=\frac{{P}_{{A}_{2}}}{{{P}^{2}}_{A}}\phantom{\rule{0ex}{0ex}}3.72=\frac{1.5}{1.5-p}\phantom{\rule{0ex}{0ex}}p=1.097atm\left(dimer\right)$

Regards.

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