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The reaction,
$$
2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons 3 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})
$$
is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \mathrm{M}$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25 \mathrm{M}$. The value for the equilibrium constant for this reaction is given by the expression
Options:
$$
2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightleftharpoons 3 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})
$$
is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \mathrm{M}$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25 \mathrm{M}$. The value for the equilibrium constant for this reaction is given by the expression
Solution:
1327 Upvotes
Verified Answer
The correct answer is:
$\left[(0.75)^{3}(0.25)\right] \div\left[(0.50)^{2}(0.75)\right]$
Thereaction-
$\quad 2 A(\mathrm{~g})+B(\mathrm{~g}) \rightleftharpoons 3 C(\mathrm{~g})+D(\mathrm{~g})$
Initial $\quad 1 \quad 1$
At equil $1-0.501-0.25 \quad 0.75 \quad 0$
$\mathrm{~K}=\frac{(0.75)^{3}(0.25)}{(0.50)^{2}(0.75)}$
$\quad 2 A(\mathrm{~g})+B(\mathrm{~g}) \rightleftharpoons 3 C(\mathrm{~g})+D(\mathrm{~g})$
Initial $\quad 1 \quad 1$
At equil $1-0.501-0.25 \quad 0.75 \quad 0$
$\mathrm{~K}=\frac{(0.75)^{3}(0.25)}{(0.50)^{2}(0.75)}$
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