nfar (abreta Quiz 6 1. You dump 5.000 grams of NH.CI solid into a 1.00 L evacuat
ID: 1027966 • Letter: N
Question
nfar (abreta Quiz 6 1. You dump 5.000 grams of NH.CI solid into a 1.00 L evacuated container at 298 You allow the reaction to reach equilibrium, and you find the total pressure in the container is 483 mmHg. NFh(g) +HCI(g) NH.cl(s) a) What is Kp for the reaction, as written above, at 298 K? b) How many grams of NH.CI are left at equilibrium in part a? NHC 2. The following reaction is at equilibrium: CaCOs(s) COdg) + Ca0(s) AH-178 klinol Decide whether the reaction is going to shift right or left if we make the following changes. a) remove all the CaO solid b) remove some of the CaO solid c) raise the temperature ght d) add CO; gas e) add N2 gas f) add H2O(I) where H,01)-co2(g) HCo,-nant 9) move to smaller V- Le-ftExplanation / Answer
We know that, Total Pressure at equillibrium= P = PNH4CL + PHCl +PNH3 .
Also, PV=nRT, where n is the total number of moles remaining at equillibrium.
For the reaction, at equillibirum assume that x grams of the added NH4Cl has reacted to give x grams of NH3 and HCl respectively.
So, remaining grams of NH4Cl is (5-x).
Also, since PV=nRT, we find that, given P is 483mm of Hg or 0.635 atm of pressure is there, we know the value of R, T is 298K and V is 1L. So, n can be found easily.
The value of n turns out to be 0.25.
The value of n will be equal to sum of individual moles of the compound. Given, their Molecular Weights and masses at equillibrium as 5-x, x and x respectively, the value of x can be found by:
n= nA + nB + nC ; where ni is the number of moles of the ith compound.
So, x turns out to be 2.3grams.
Now, Kp = (Partial pressure of NH4Cl)/ (Partial pressure of HCl)*(Partial pressure of NH3)
= (mole fraction of NH4Cl *P)/ (mole fraction of HCl * P)*(mole fraction of NH3 * P)
Putting the values, we get Kp as 9.34.
(b.) Also, the remaining grams of NH4Cl at eqm is 5-x, which is (5-2.3)g or 2.7grams.
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