At 500K, NOCl is 90% decomposed for the reaction 2NOCl<->2NO+Cl2. If we assume that 1.00mol of NOCl in 1.00L before dissociation, what is the value of Kc for equilibrium at 500K?

Question

At 500K, NOCl is 90% decomposed for the reaction 2NOCl<->2NO+Cl2. If we assume that 1.00mol of NOCl in 1.00L before dissociation, what is the value of Kc for equilibrium at 500K?

1 Answer

Impact of this question

1450 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-11T15:49:29

$K_{c} ≅ 36$ $K_c~=36$ ....

Explanation:

For $2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)$ $2NOCl(g) rightleftharpoons2NO(g) + Cl_2(g)$

We write $K_{c} = \frac{{[N O]}^{2} [C l_{2}]}{{[N O C l (g)]}^{2}}$ $K_c=([NO]^2[Cl_2])/([NOCl(g)]^2)$

Now initially there were $1.00 \cdot m o l \cdot L^{- 1}$ $1.00*mol*L^-1$ with respect to $N O C l$ $NOCl$ ; at equilibrium $[N O C l] = 0.10 \cdot m o l \cdot L^{- 1}$ $[NOCl]=0.10*mol*L^-1$ , $[N O] = 0.90 \cdot m o l \cdot L^{- 1}$ $[NO]=0.90*mol*L^-1$ , and $[C l_{2}] = 0.45 \cdot m o l \cdot L^{- 1}$ $[Cl_2]=0.45*mol*L^-1$ ...

And so we fill in the numbers....

$K_{c} = \frac{{(0.90 \cdot m o l \cdot L^{- 1})}^{2} \times 0.45 \cdot m o l \cdot L^{- 1}}{{(0.10)}^{2}} = 36.5$ $K_c=((0.90*mol*L^-1)^2xx0.45*mol*L^-1)/(0.10)^2=36.5$

Science

Math

Humanities

... and beyond

At 500K, NOCl is 90% decomposed for the reaction 2NOCl<->2NO+Cl2. If we assume that 1.00mol of NOCl in 1.00L before dissociation, what is the value of Kc for equilibrium at 500K?

1 Answer

Explanation:

Related questions

Impact of this question