Chemistry, 15.10.2019 17:30 flargenshpargle7423
At 900.0 k, the equilibrium constant (kp) for the following reaction is 0.345. 2so2 + o2(g) → 2so3(g) at equilibrium, the partial pressure of so2 is 35.0 atm and that of o2 is 15.9 atm. the partial pressure of so3 is atm. at 900.0 k, the equilibrium constant (kp) for the following reaction is 0.345. 2so2 + o2(g) → 2so3(g) at equilibrium, the partial pressure of so2 is 35.0 atm and that of o2 is 15.9 atm. the partial pressure of so3 is atm. 192 6.20 × 10-4 4.21 × 10-3 40.2 82.0
Answers: 3
![Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm](/tpl/images/0322/3226/67e1c.png)
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At 900.0 k, the equilibrium constant (kp) for the following reaction is 0.345. 2so2 + o2(g) → 2so3(g...
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