Consider the following processes Δ H(~kJ / mol) array ll 1 / 2 A arrow B & +150 3 B arrow 2 C+D & -125 E+A arrow 2 D & +350 array For B+D arrow E+2 C, Δ H will be

Correct answer: B. -175 ~kJ / mol aligned & (1)/(2) A B ; Δ H=150 kJ / mol & 3 B 2 C+D ; Δ H=-125 kJ / mol & E+A 2 D ; Δ H=+350 kJ / mol & By [2 ×(i)+(ii)]-(iii), we have & B+D E+2 C & Δ H=150 × 2+(-125)-350 & =-175 kJ / mol . aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2011 · Chemistry · Thermodynamics (C)

Consider the following processes Δ H(~kJ / mol) array ll 1 / 2 A...

Consider the following processes

\Delta H(\mathrm{~kJ} / \mathrm{mol})

\begin{array}{ll}1 / 2 A \rightarrow B & +150 \\ 3 B \rightarrow 2 C+D & -125 \\ E+A \rightarrow 2 D & +350\end{array}

For

B+D \rightarrow E+2 C, \Delta H

will be

Options

A $525 \mathrm{~kJ} / \mathrm{mol}$
B $-175 \mathrm{~kJ} / \mathrm{mol}$
C $-325 \mathrm{~kJ} / \mathrm{mol}$
D $325 \mathrm{~kJ} / \mathrm{mol}$

Correct answer

B. $-175 \mathrm{~kJ} / \mathrm{mol}$

Step-by-step solution

\begin{aligned} & \frac{1}{2} A \longrightarrow B ; \quad \Delta H=150 \mathrm{ kJ} / \mathrm{mol} \\ & 3 B \longrightarrow 2 C+D ; \Delta H=-125 \mathrm{ kJ} / \mathrm{mol} \\ & E+A \longrightarrow 2 D ; \Delta H=+350 \mathrm{ kJ} / \mathrm{mol} \\ & \hline \text{ By }[2 \times(\mathrm{i})+(\mathrm{ii})]-(\mathrm{iii}), \text{ we have } \\ & B+D \longrightarrow E+2 C \\ & \therefore \quad \Delta H=150 \times 2+(-125)-350 \\ & \quad=-175 \mathrm{ kJ} / \mathrm{mol} . \end{aligned}

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