Kp=Kc[RT]n
Kc=Kp/[RT]n
n= moles of gaseous product - moles of gaseous reactant = 3-2=1
625oc=898K
ie: Kc= 0.269/(0.08206 x 898)1
Kc=0.00365
2) For the equilibrium: 2 SO3(g) <=> O2(g) + 2 SO2(g) Kp = 0.269 at 625...
2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = K[R R = 0.08206 L-atm/mol K (5pts)
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at 625 oC What is Kc at this temperature? Kp = Kc[RT]Δn R = 0.08206 L-atm/mol K
2) For the equilibrium: 2 SO2(g) + O2(g) < => 2 503(g) Kp = 2.98 at 875oC What is Ke at this temperature? Ko-K[RT]An R = 0.08206 L-atm/mol K (5pts)
5. For the reaction... SO2 + O2 <=> SO3 If the equilibrium position shifts to the right, the concentration of O2 will (2 points) o remain constant increase Decrease
how to solve this? Sulphur trioxide decomposes at high temperature in a sealed container: 2 SO3(g) <--> 2 SO2(g) + O2(g). Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K to 3 decimal places.
For the equilibrium 2 SO3(g) <----> 2 SO2(g) + O2(g), Kc is 4.08 * 10-3 at 1000 K. Calculate the value for Kp.
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Given the reaction: 2 SO2(g) + O2(g) ⇌ 2 SO3(g) Starting with 0.100 M of SO2 and 0.050 M of O2, which value of K should be used and why? a. KC because the chemicals are in the gas phase. b. KP because the concentrations are in Molarity. c. KP because the chemicals are in the gas phase. d. KC because the concentrations are in Molarity.
A reaction vessel contains an equilibrium mixture of SO2, O2,, and SO3. The reaction proceeds such that: 2SO2 (g) + O2 (g) ----> <---- 2SO3 (g) The partial pressures at equilibrium are Pso2=0.001111 atm PO2 = 0.002728 atm PSO3 =0.0166 Calculate the Kp for the reaction.
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