MCQ
What will be the final major product ?
- A
- B
- ✓
- D
✓
Answer
Correct option: C.
c
Ring containing sulphur is more strained.
Ring containing sulphur is more strained.
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→Given below are two statements:
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In the light of the above statements, choose the correct answer from the options given below:
For the reaction:
→$RCH _2 Br + I ^{-} \stackrel{\text { Acetone }}{\longrightarrow} \underset{\text { major }}{ RCH _2 I + Br ^{-}}$
The correct statement is :
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Consider the following reaction
→${C_6}{H_6}(l)\, + \,\frac{{15}}{2}\,{O_2}(g)\, \to \,6\,C{O_2}(g)\, + \,3{H_2}O\,(g)$
Signs of $\Delta H,\,\Delta S$ and $\Delta G$ for the above reaction will be
Given
$
\begin{array}{l}
C_{\text {graphite }}+O_2 \rightarrow CO_2(g): \\
\Delta_{r} H^0=-393.5 KJ / mol
\end{array}
$
$
H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l)
$
$
\Delta_r H^0=-285.8 kJ / mol
$
$
CO_2(g)+2 H_2 O(l) \rightarrow CH_4(g)+2 O_2(g):
$
$
\Delta_r H^0=+890.3 kJ mol^{-1}
$
Based on the above thermochemical equations, the value of $\Delta_r H^0$ at 298 K for the reaction
$
C_{\text {(graphite) }}+2 H_2(g) \rightarrow CH_4(g) \text { will be : }
$
→$
\begin{array}{l}
C_{\text {graphite }}+O_2 \rightarrow CO_2(g): \\
\Delta_{r} H^0=-393.5 KJ / mol
\end{array}
$
$
H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l)
$
$
\Delta_r H^0=-285.8 kJ / mol
$
$
CO_2(g)+2 H_2 O(l) \rightarrow CH_4(g)+2 O_2(g):
$
$
\Delta_r H^0=+890.3 kJ mol^{-1}
$
Based on the above thermochemical equations, the value of $\Delta_r H^0$ at 298 K for the reaction
$
C_{\text {(graphite) }}+2 H_2(g) \rightarrow CH_4(g) \text { will be : }
$
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→