MCQ
The equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) shifts forward if:
- ACatalyst is used.
- BAn adsorbent is used to remove SO3 as soon as it is formed.
- CLarge amount of products are used.
- DSmall amount of reactants are used.
Explanation:
Removal of any product or adding of reactants favours forward reaction i.e., SO3 formation. This is according to Le- Chatelier's principle.
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|
List-$I$ (Molecule) |
List-$II$. (Number and types of bond/s between two carbon atoms) |
| $A$. ethane | $I$. one $\sigma$-bond and two $\pi$-bonds |
| $B$. ethene | $II$. two $\pi$-bonds |
| $C$. carbon molecule, $\mathrm{C}_2$ | $III$. one $\sigma$-bond |
| $D$. ethyne | $IV$. one $\sigma$-bond and one $\pi$-bond |
Choose the correct answer from the options given below:
$\Delta_f G^0[\mathrm{C}(\text { graphite })]=0 \mathrm{kJmol}^{-1}$
$\Delta_f G^0[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJmol}^{-1}$
The standard state means that the pressure should be $1$ bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mol}^{-1}$. If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $\mathrm{T}=298 \mathrm{~K}$, the pressure at which $C$ (graphite) is in equilibrium with C(diamond), is
[Useful information: $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^5 \mathrm{~Pa}$ ]
Statement $I$ : The boiling point of three isomeric pentanes follows the order
n-pentane $>$ isopentane $>$ neopentane
Statement $II$ : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point.
In the light of the above statements, choose the most appropriate answer from the options given below: