A substance X in presence of OH- reacts by two different mechanisms to give two different isomeric products as shownIf rate law is given as −d[X]dt=k1[X]OH−+k2[X] and ΔHreaction-I =−40kJ and ΔHreaction-II =−28kJ then calculate |ΔH| when 1 mole of X completely reacts [in kJ] if[OH-] is maintained at 0.1 M.

Rate of decomposition,−d[X]dt=k1×0.1×[X]+k2[X]=4×10−3×[X]k1×0.1k2=y1−y⇒ y=141−y=34Change in enthalpy,ΔH=−14×40+34×28=−31

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A substance X in presence of OH^- reacts by two different mechanisms to give two different isomeric products as shown
If rate law is given as $\frac{- d [X]}{d t} = k_{1} [X] [{OH}^{-}] + k_{2} [X]$ and $Δ H_{reaction-I} = - 40 kJ$ and $Δ H_{reaction-II} = - 28 kJ$ then calculate $| Δ H |$ when 1 mole of X completely reacts [in kJ] if[OH^-] is maintained at 0.1 M.

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answer is 31.

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Detailed Solution

Rate of decomposition,

$- \frac{d [X]}{d t} = k_{1} \times 0.1 \times [X] + k_{2} [X] = 4 \times 10^{- 3} \times [X]$

$\frac{k_{1} \times 0.1}{k_{2}} = \frac{y}{1 - y}$

$\begin{array}{l} \Rightarrow y = \frac{1}{4} \\ 1 - y = \frac{3}{4} \end{array}$

Change in enthalpy,

$Δ H = - [\frac{1}{4} \times 40 + \frac{3}{4} \times 28] = - 31$

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