At 603 K, Kc for the reaction NH4Cl <===> NH3 + HCl is 0.011. What is the equilibrium concentration of NH3 if the initial concentration of NH4Cl is 0.500 M. Please help me with this question because i got 2 answers so prove me.
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Why not show your work for the two answers you have so we can find the error. That's a lot easier than us working the problem and letting you look for the error.
......NH4Cl(g) --> NH3(g) + HCl(g)
I......0.5..........0........0
C......-x...........x........x
E.....0.5-x.........x........x
Kc = 0.011 = (x)(x)/(0.50x)
Solve for x.
......NH4Cl(g) --> NH3(g) + HCl(g)
I......0.5..........0........0
C......-x...........x........x
E.....0.5-x.........x........x
Kc = 0.011 = (x)(x)/(0.50x)
Solve for x.
Please answer me with the working out...
please can you help me with the correct answer
0 answers