Consider the following reaction:
2NOCl(g) 2NO(g) + Cl2(g)
Initially pure NOCl(g) is placed in a vessel at 2.96 atm. At equilibrium, 0.410% of the NOCl has decomposed. Determine the
value for Kp.
Can I get an asnwer for this?
*answer
To determine the value of Kp for the given reaction, we need to first calculate the partial pressures of each species at equilibrium.
Let's consider that x is the partial pressure of Cl2(g) at equilibrium. Since 2 moles of Cl2 are formed for every 2 moles of NOCl decomposed, the partial pressure of Cl2(g) can be expressed as 2x.
The partial pressure of NO(g) at equilibrium can be calculated using the initial pressure of NOCl and the extent of the reaction. As 0.410% of NOCl has decomposed, it means that 0.410/100 = 0.0041 moles of NOCl is decomposed. Thus, the number of moles of NO(g) formed is also 0.0041 moles. Taking into account the stoichiometry, the partial pressure of NO(g) can be expressed as 0.0041/atm.
Now we can construct the expression for Kp by substituting the partial pressures into the equilibrium expression:
Kp = (P(NO)^2 * P(Cl2)) / P(NOCl)^2
Considering all the values we calculated, we have:
Kp = ((0.0041/atm)^2 * (2x)) / (2.96 atm - 0.0041 atm)^2
Simplifying further, we get:
Kp = (0.0041^2 * (2x)) / (2.9569)^2
Finally, we need to determine the value of x. Since the decomposition of NOCl reaches equilibrium, the total pressure inside the vessel remains constant. Therefore, we can set up an expression to find x:
2x + (0.0041/atm) = 2.96 atm
Solving for x, we get:
2x = 2.96 atm - 0.0041 atm
x = 2.9559 / 2
Now we can substitute this value of x back into the expression for Kp:
Kp = (0.0041^2 * (2 * 2.9559 / 2)) / (2.9569)^2
Evaluating this expression will give us the value of Kp for the given reaction.