NEET 2010
- Exam
- NEET
- Year
- 2010
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NEET 2010 · Chemistry · Chemical Equilibrium
The reaction, (2 ~A_ (g) +B_ (g) rightharpoons 3 C_ (g) +D_ (g) )...
The reaction,
(2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \leftrightharpoons 3 \mathrm{C}_{(\mathrm{g})}+\mathrm{D}_{(\mathrm{g})})
is begun with the concentrations of and both at an initial value of 1.00 M When equilibrium is reached, the concentration of is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
Options
- A(\left[(0.75)^3(0.25)\right] \div\left[(1.00)^2(1.00)\right])
- B(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right])
- C(\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.25)\right])
- D(\left[(0.75)^3(0.25)\right] \div\left[(0.75)^2(0.25)\right])
Correct answer
B. (\left[(0.75)^3(0.25)\right] \div\left[(0.50)^2(0.75)\right])
Step-by-step solution
()
Equilibrium constant, (K=\frac{[C]^3[D]}{[A]^2[B]})
(\therefore \quad K=\frac{(0.75)^3(0.25)}{(0.5)^2(0.75)})