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A hypothetical reaction $A \longrightarrow 2 B$ proceeds through the following sequence of steps
I. $A \longrightarrow C ; \Delta_r H=q_1$
II. $C \longrightarrow D ; \Delta_r H=q_2$
III. $\frac{1}{2} D \longrightarrow B ; \Delta_r H=q_3$
The heat of hypothetical reaction is
Options:
I. $A \longrightarrow C ; \Delta_r H=q_1$
II. $C \longrightarrow D ; \Delta_r H=q_2$
III. $\frac{1}{2} D \longrightarrow B ; \Delta_r H=q_3$
The heat of hypothetical reaction is
Solution:
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Verified Answer
The correct answer is:
$q_1+q_2+2 q_3$
One mole of $A$ as reactant So, we write down equation
(i) Two moles of $B$ as product, so we multiply equation (iii) by (b). Finally, we add the three equation term-by-term
$\begin{array}{ll}A \longrightarrow C & q_1 \\ C \longrightarrow D & q_2 \\ D \longrightarrow 2 B & 2 q_3 \\ A \longrightarrow 2 B & q_1+q_2+2 q_3\end{array}$
Therefore, the heat of hypothetical reaction is $q_1+q_2+2 q_3$.
(i) Two moles of $B$ as product, so we multiply equation (iii) by (b). Finally, we add the three equation term-by-term
$\begin{array}{ll}A \longrightarrow C & q_1 \\ C \longrightarrow D & q_2 \\ D \longrightarrow 2 B & 2 q_3 \\ A \longrightarrow 2 B & q_1+q_2+2 q_3\end{array}$
Therefore, the heat of hypothetical reaction is $q_1+q_2+2 q_3$.
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