Download App

5. Thermodynamics and thermochemistry — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry5. Thermodynamics and thermochemistry1 Mark
MCQ
For which of the following processes, $\Delta S$  is negative?
  • A
    $C$ (diamond) $\to$  $C$ (graphite)
  • B
    $N_2(g,\,1\,atm) \to  N_2(g,\,5\,atm)$
  • C
    $N_2(g,\,273\,K) \to  N_2(g,\,300\,K)$
  • D
    $H_2(g) \to  2H(g)$
✓

Answer

$(a)$ When diamond is converted into graphite when it is heated to $1500\,^oC$ and entropy is increased so $\Delta S > 0$
$(b)$ When pressure increases then molecules of gas will come closer and intermolecular distance decreases so entropy will also decrease and $\Delta S < 0$
$(c)$ When we increase the temperature of a gas then random ness is increased due to kinetic energy gained by molecules. So, $\Delta S > 0$
$(d)$  $H_2$ molecule is converted into atoms, the no. of particles increases . Thus entropy will increase. So, $\Delta S > 0$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 5. Thermodynamics and thermochemistry→5. Thermodynamics and thermochemistry — all question types→Chemistry STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The dissociation constant of an acid $HA$  is $1 \times {10^{ - 5}}$. The $pH$ of $0.1$ molar solution of the acid will be
→
On adding $0.1\, M$ solution each of $[Ag^+], [Ba^{2+}], [Ca^{2+}]$ in a $Na_2SO_4$ solution, species first precipitated is $[K_{sp}\, BaSO_4 = 10^{-11}, K_{sp}\, CaSO_4 = 10^{-6}, K_{sp}\,Ag_2SO_4 = 10^{-5}]$
→
Number of electrons in the valence orbit of nitrogen in an ammonia molecule are
→
When ethyl iodide and propyl iodide react with $Na$ in the presence of ether, they form
→
What is the equilibrium expression for the reaction ${P_{4(s)}} + 5{O_{2(g)}}$ $\rightleftharpoons$ ${P_4}{O_{10(s)}}$
→
The value of $n$ in the molecular formula $\mathrm{Be_n} \mathrm{Al}_2 \mathrm{Si}_6 \mathrm{O}_{18}$ is :
→
$IUPAC $ name of $C{H_3} - CH = CH - COOH$
→
The single Bond covalent radius $(SBCR)$ of $P$ is $0.11\,nm.$ The single Bond covalent radius of $Cl$ is : ( $P =$ phosphorus)
→
The elements having the electronic configuration $[Kr]4d^{10}\,f^{14}\,5s^2\,p^6\,d^2\,6s^2$ belongs to
→
In the reaction :

${C_6}{H_5} - H\mathop {\xrightarrow{{[X]}}}\limits_{AlC{l_3}} $  $\begin{matrix}
   O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   ||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   {{C}_{6}}{{H}_{5}}-C-C{{H}_{2}}-{{C}_{6}}{{H}_{5}}  \\
\end{matrix}$

$[X]$ will be :

→