A) 0.039
B) 0.078
C) 2
D) 5
E) None of the above
F)
Correct Answer: E
Solution :
\[2G\,\,\,\,\,\,\,\,\,\,\,+J\,\,\,\,\,\,\,\,\,\,\,\,\,\,D\,\,\,\,\,\,\,+2T\] Att = 01 mol 2mol 0.5 mol 0 mol At equil. \[(1-2x)\]\[(2-x)\] \[(0.5+x)\] \[2x\] \[{{K}_{c}}=\frac{[D]\,\,[T]}{{{[G]}^{2}}\,\,[J]}\] \[1.5\times {{10}^{3}}=\frac{(0.5+x)\times 2\,x}{(1-2x)\,(2-x)}\] \[1500=\frac{x+2{{x}^{2}}}{2{{x}^{2}}-5x+2}\] \[\therefore \] \[x+2{{x}^{2}}=3000{{x}^{2}}-7500x+3000\] or \[2998{{x}^{2}}-7499x+3000=0\] or \[2{{x}^{2}}-5x+2=0\] (On dividing both sides from 1500) or \[(x-2)\,\,(2x-1)=0\] If \[x=2=0\Rightarrow x=2\] If \[2x-1=0\Rightarrow x=\frac{1}{2}=0.5\] \[\therefore \] Concentration of T at equilibrium \[=2x\] = 2 or 1
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