Question

For the reaction below, Kp = 1.59 at 100°C. If 1.0 g of SrCO3 is placed in an empty 5.00 L reactor and allowed to reach equilibrium, what will be the pressure in the reactor?

Answer 1

Answer:

$p_{CO_2}^{eq}=1.59atm$

Explanation:

Hello, in this case, one could consider the undergoing chemical reaction as:

$SrCO_3(s)\rightleftharpoons SrO(s)+CO_2(g)$

Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:

$Kp=p_{CO_2}^{eq}=1.59atm$

Hence, since at the beginning there is no carbon dioxide, its pressure at equilibrium equals Kp:

$Kp=p_{CO_2}^{eq}=1.59atm$

Which was clearly defined above.

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