At 25.0 degree celcius, the equilibrium 2 NOBr (gas) --> 2 NO (gas) + BR2 (gas) is rapidly established. When 1.10 g of NOBr is placed in a 1.0 L vessel at 25.0 degree celcius, the equilibrium pressure is 0.355 bar. Calculate the equilibrium constant K. (Partial pressure P = 1 bar )
Sorry the last part partial pressure P = 1 bar is suppose to be standard pressure = 1 bar )
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To calculate the equilibrium constant K for the given reaction, we first need to determine the partial pressures of NO and Br2 at equilibrium.
Given:
Initial pressure of NOBr = 0.355 bar
Initial volume = 1.0 L
Since the reaction is rapidly established, we can assume that the change in volume is negligible. Therefore, the final volume is also 1.0 L.
At equilibrium, the reaction would have proceeded as follows:
2 NOBr (g) -> 2 NO (g) + Br2 (g)
Let's assume that at equilibrium, the partial pressure of NOBr is P(NOBr), the partial pressure of NO is P(NO), and the partial pressure of Br2 is P(Br2).
Since 1.10 g of NOBr is placed in a 1.0 L vessel, we can determine the initial moles (n) of NOBr using the ideal gas law. The molar mass of NOBr is 109.0 g/mol.
n(NOBr) = mass / molar mass
n(NOBr) = 1.10 g / 109.0 g/mol
Next, we use the ideal gas law to calculate the initial pressure of NOBr:
PV = nRT
P(NOBr) = n(NOBr) * R * T / V
P(NOBr) = (1.10 g / 109.0 g/mol) * (0.0831 L bar / mol K) * (298 K) / 1.0 L
Calculate P(NO) and P(Br2) at equilibrium:
P(NO) = 2 * P(NOBr)
P(Br2) = P(NOBr)
Now, we can calculate the equilibrium constant K using the formula:
K = (P(NO)^2 * P(Br2)) / P(NOBr)^2
Substitute the values we calculated into the formula and solve for K.
Note: In the original question, it is mentioned that partial pressure P = 1 bar, but partial pressure should not have been given as 1 bar. Therefore, please ignore the provided partial pressure and follow the steps explained above for the accurate calculation of the equilibrium constant K.