A) 1
B) 10
C) 5
D) 0.33
Correct Answer: A
Solution :
\[{{H}_{2}}\ +\ {{I}_{2}}\] ⇌ 2HI Initial conc. 4.5 4.5 0 x x 2x from question 2x = 3 \[x=\frac{3}{2}=1.5\] So conc. at eqm. \[4.5-1.5\] of \[{{H}_{2}}\] \[=4.5-1.5\] of \[{{I}_{2}}\] and 3 of HI \[K=\frac{{{[HI]}^{2}}}{[{{I}_{2}}]\,\,[{{H}_{2}}]}=\frac{3\times 3}{3\times 3}=1\].You need to login to perform this action.
You will be redirected in
3 sec