Question

The reaction ${ 2A_{(g)} + B_{(g)} <=> 3C_{(g)} + D_{(g)}}$ is begun with the concentrations of $A$ and $B$ both at an intial value of $1.00\, M$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25 \,M$. The value for the equilibrium constant for this reaction is given by the expression.

• A. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)]$
• B. $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.25)]$
• C. $[(0.75)^3 (0.25)] \div [(0.75)^2 (0.25)]$
• D. $[(0.75)^3 (0.25)] \div [(1.00)^2 (1.00)]$

Answer 1

Answer: (A)

$[(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)]$

Answer 2

The correct option is(A): $[(0.75)^3 (0.25)] \div [(0.50)^2 (0.75)]$

Given,
(I)$F{{e}_{2}}O(s)+3CO(g)\xrightarrow[{}]{{}}2Fe(s)+3C{{O}_{2}}(g);$
$\Delta H=-26.8\,\text{kJ}$
(II)$FeO(s)+CO(g)\xrightarrow[{}]{{}}Fe(s)+C{{O}_{2}}(g);$
$\Delta H=-16.5\,\text{kJ}$
On multiplying Eq (II) with 2, we get
(III) $2FeO(s)+2CO(g)\xrightarrow[{}]{{}}2Fe(s)+2C{{O}_{2}}(g);$
$\Delta H=-33\,\text{kJ}$
On subtracting Eq (III) from I, we get
$F{{e}_{2}}{{O}_{3}}(s)+CO(g)\xrightarrow[{}]{{}}2FeO(s)+C{{O}_{2}}(g);$
$\Delta H=-26.8-(-33)$
$=+6.2\,\text{kJ}$