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SOVA2 [1]
3 years ago
8

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 equilibr

ium, what will be the pressure in the reactor?
Chemistry
1 answer:
Sergio [31]3 years ago
8 0

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.

Best regards.

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