Looking for classes? Ksquare Career Institute, Bengaluru →
Chemistry
Start Practice Clear paper filter
Showing 1–25 of 29 questions
-
Identify the correct statement for change of Gibbs energy for a system ($\Delta G_{system}$) at constant temperature and pressure :-At constant temperature and pressure, the Gibbs energy change $\Delta G = \Delta H - T\Delta S$ is the criterion of spontaneity.
If $\Delta G_{system} \lt 0$, the process is spontaneous.
If $\Delta G_{system} \gt 0$, the process is non-spontaneous (the reverse process is spontaneous).
If $\Delta G_{system} = 0$, the process is in equilibrium; there is no net change in either direction.
So the statements saying that $\Delta G \gt 0$ is spontaneous, that $\Delta G \lt 0$ is not spontaneous, and that at $\Delta G = 0$ the system is still moving in a particular direction are all wrong.
The correct statement is: if $\Delta G_{system} = 0$, the system has attained equilibrium. -
A plot of $\log\dfrac{x}{m}$ versus $\log p$ for the adsorption of a gas on a solid gives a straight line with slope equal to :Freundlich adsorption isotherm: $\dfrac{x}{m} = K\,p^{1/n}$ (with $n \gt 1$)
Taking logarithms on both sides: $\log\dfrac{x}{m} = \log K + \dfrac{1}{n}\log p$
This is of the form $y = c + mx$, the equation of a straight line.
So a plot of $\log\dfrac{x}{m}$ against $\log p$ is a straight line with slope $= \dfrac{1}{n}$
and intercept on the y-axis $= \log K$.
Hence the slope is equal to $\dfrac{1}{n}$. -
Assume each reaction is carried out in an open container. For which reaction will $\Delta H = \Delta E$ ?$\Delta H = \Delta E + \Delta n_g RT$, where $\Delta n_g$ = moles of gaseous products $-$ moles of gaseous reactants.
$\Delta H = \Delta E$ only when $\Delta n_g = 0$.
$H_2(g) + Br_2(g) \rightarrow 2HBr(g)$: $\Delta n_g = 2 - 2 = 0$, so $\Delta H = \Delta E$.
$C(s) + 2H_2O(g) \rightarrow 2H_2(g) + CO_2(g)$: $\Delta n_g = 3 - 2 = +1$ (the solid carbon is not counted).
$PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)$: $\Delta n_g = 2 - 1 = +1$.
$2CO(g) + O_2(g) \rightarrow 2CO_2(g)$: $\Delta n_g = 2 - 3 = -1$.
So $\Delta H = \Delta E$ for $H_2(g) + Br_2(g) \rightarrow 2HBr(g)$. -
During the process of digestion, the proteins present in food materials are hydrolysed to amino acids. The two enzymes involved in the process:
Proteins $\xrightarrow{\text{Enzyme (A)}}$ Polypeptides $\xrightarrow{\text{Enzyme (B)}}$ Amino acids,
are respectively -Proteins are polymers of amino acids joined by peptide bonds; enzymes that hydrolyse peptide bonds are called proteolytic enzymes.
Pepsin, present in the gastric juice of the stomach, hydrolyses proteins into smaller polypeptides (proteoses and peptones).
Trypsin, present in the pancreatic juice, acts in the small intestine and breaks the polypeptides down further, finally giving amino acids.
Amylase and diastase act on starch, and maltase acts on maltose.
Lipase acts on fats.
Invertase hydrolyses sucrose, and zymase converts glucose into ethanol.
So enzyme (A) is pepsin and enzyme (B) is trypsin. -
The human body does not produce :-Vitamins are organic compounds required in small amounts for normal growth and health.
Most vitamins cannot be synthesised by the human body, so they must be supplied in the diet; that is why they are called essential dietary factors.
(Only a few are made in limited amounts, for example vitamin D in the skin in sunlight.)
DNA is synthesised in every dividing cell by replication.
Hormones are produced by the endocrine glands.
Enzymes are proteins synthesised in the cells of the body.
So the human body does not produce vitamins. -
More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is :-Lanthanoids show mainly the +3 oxidation state, with +2 and +4 only in a few cases.
Actinoids show a much wider range of oxidation states, from +3 up to +7 (for example U shows +3, +4, +5, +6 and Np shows up to +7).
The reason is that in actinoids the 5f, 6d and 7s orbitals have comparable energies; the energy difference between 5f and 6d is very small.
So electrons from all three kinds of orbitals can take part in bonding.
In lanthanoids the energy gap between the 4f and 5d orbitals is larger, and the 4f electrons are deeply buried, so they are not easily available for bonding.
Hence the main reason is the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals. -
In which of the following pairs are both the ions coloured in aqueous solution -
[At. No. : Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27]Transition metal ions are coloured when they have unpaired electrons in partially filled d orbitals, so that d-d transitions can absorb visible light.
Ions with $d^0$ or $d^{10}$ configuration have no d-d transitions and are colourless.
$Sc^{3+}$: $3d^0$, colourless.
$Ti^{3+}$: $3d^1$, one unpaired electron, coloured (purple).
$Ni^{2+}$: $3d^8$, two unpaired electrons, coloured (green).
$Co^{2+}$: $3d^7$, three unpaired electrons, coloured (pink).
$Cu^+$: $3d^{10}$, colourless.
Every pair containing $Sc^{3+}$ or $Cu^+$ has one colourless ion, so the pair in which both ions are coloured is $Ni^{2+}$ and $Ti^{3+}$. -
The enthalpy and entropy change for the reaction:
$Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$
are 30 kJ mol$^{-1}$ and 105 J K$^{-1}$ mol$^{-1}$ respectively. The temperature at which the reaction will be in equilibrium is :-For the reaction $Br_2(l) + Cl_2(g) \rightarrow 2BrCl(g)$: $\Delta H = 30$ kJ mol$^{-1}$ and $\Delta S = 105$ J K$^{-1}$ mol$^{-1}$
At equilibrium, $\Delta G = 0$
$\Delta G = \Delta H - T\Delta S = 0$
$\therefore \Delta H = T\Delta S$
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{30 \times 1000}{105}$ (converting kJ to J)
$T = 285.7$ K -
The appearance of colour in solid alkali metal halides is generally due to :When an alkali metal halide crystal is heated in the vapour of the alkali metal, metal atoms deposit on the surface.
Halide ions diffuse to the surface and combine with these atoms, and the electrons released by the ionising metal atoms diffuse into the crystal and occupy the vacant anion sites.
The anion vacancies occupied by unpaired electrons are called F-centres (from the German word Farbenzenter, meaning colour centre).
These electrons absorb energy from visible light and get excited, which gives colour to the crystal.
Because of the F-centre defect a colourless ionic crystal is converted into a coloured one: NaCl becomes yellow, KCl violet (lilac) and LiCl pink.
Schottky and Frenkel defects are stoichiometric defects and do not produce colour.
So the colour is generally due to F-centres. -
The general molecular formula, which represents the homologous series of alkanols is :-Alkanols are saturated monohydric alcohols, obtained by replacing one hydrogen atom of an alkane by an OH group.
General formula of an alkane: $C_nH_{2n+2}$
General formula of an alkanol: $C_nH_{2n+1}OH$
Written as a molecular formula this is $C_nH_{2n+2}O$.
Examples: methanol $CH_4O$ ($n = 1$), ethanol $C_2H_6O$ ($n = 2$).
$C_nH_{2n}O$ represents aldehydes and ketones, and $C_nH_{2n}O_2$ represents carboxylic acids and esters.
So the general molecular formula of alkanols is $C_nH_{2n+2}O$. -
If $E^\circ_{Fe^{2+}/Fe} = -0.441$ V and $E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771$ V, the standard EMF of the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$ will be :For the cell reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$:
Anode reaction (oxidation): $Fe \rightarrow Fe^{2+} + 2e^-$
Cathode reaction (reduction): $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$, where both $E^\circ$ values are reduction potentials.
$E^\circ_{cell} = 0.771 - (-0.441)$
$E^\circ_{cell} = 1.212$ V
EMF is an intensive property, so the potentials are not multiplied by the number of electrons. -
For the reaction :
$2A + B \rightarrow 3C + D$
Which of the following does not express the reaction rate :-For the reaction $2A + B \rightarrow 3C + D$, the rate is expressed as:
Rate $= -\dfrac{1}{2}\dfrac{d[A]}{dt} = -\dfrac{d[B]}{dt} = +\dfrac{1}{3}\dfrac{d[C]}{dt} = +\dfrac{d[D]}{dt}$
The concentration of a reactant decreases with time, so a negative sign is used to make the rate positive.
The concentration of a product increases with time, so a positive sign is used.
C is a product, so the correct expression for it is $+\dfrac{d[C]}{3\,dt}$.
$-\dfrac{d[C]}{3\,dt}$ would be a negative quantity, so it does not express the reaction rate. -
$[NH(CH_2)_6NHCO(CH_2)_4CO]_n$ is a :-The given polymer is Nylon-6,6.
It is formed from two different monomers: hexamethylenediamine, $H_2N(CH_2)_6NH_2$, and adipic acid, $HOOC(CH_2)_4COOH$.
$n\,H_2N(CH_2)_6NH_2 + n\,HOOC(CH_2)_4COOH \rightarrow [NH(CH_2)_6NHCO(CH_2)_4CO]_n + 2n\,H_2O$
A polymer made from two or more different monomers is called a copolymer; a homopolymer is made from only one kind of monomer.
It is formed with the loss of water molecules, so it is a condensation polymer, not an addition polymer.
It is a linear polyamide that softens on heating, i.e. a thermoplastic fibre, not a thermosetting polymer.
So the polymer is a copolymer. -
Which one of the following is a peptide hormone :-Hormones are classified chemically into steroid hormones, peptide (polypeptide or protein) hormones and amine hormones.
Glucagon is a peptide hormone made of 29 amino acids; it is secreted by the $\alpha$-cells of the pancreas and raises the blood glucose level.
Other peptide hormones are insulin and oxytocin.
Testosterone is a steroid hormone.
Thyroxine and adrenaline are amine hormones, derived from the amino acid tyrosine.
So the peptide hormone is glucagon. -
Nucleophilic addition reaction will be most favoured in :-In nucleophilic addition the nucleophile attacks the electron-deficient carbonyl carbon.
Reactivity $\propto \dfrac{1}{\text{steric hindrance}}$: bulky groups around the carbonyl carbon hinder the approach of the nucleophile.
Alkyl groups are also electron releasing ($+I$ effect); they reduce the positive charge on the carbonyl carbon and lower its reactivity.
So aldehydes (one alkyl group) are more reactive than ketones (two alkyl groups), and among aldehydes the smaller alkyl group gives the higher reactivity.
$CH_3CHO$ has the smallest alkyl group among the given compounds; $CH_3CH_2CHO$ has a larger ethyl group, and the two ketones are still less reactive.
Order: $CH_3CHO \gt CH_3CH_2CHO \gt (CH_3)_2CO \gt CH_3CH_2CH_2COCH_3$
So nucleophilic addition is most favoured in $CH_3CHO$. -
Consider the reaction
$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$
The equality relationship between $\dfrac{d[NH_3]}{dt}$ and $-\dfrac{d[H_2]}{dt}$ is :-For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$, the rate is:
Rate $= -\dfrac{d[N_2]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = +\dfrac{1}{2}\dfrac{d[NH_3]}{dt}$
Taking the last two terms: $\dfrac{1}{2}\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt}$
Multiplying both sides by 2:
$\dfrac{d[NH_3]}{dt} = -\dfrac{2}{3}\dfrac{d[H_2]}{dt}$
This agrees with the stoichiometry: 3 moles of $H_2$ consumed give 2 moles of $NH_3$. -
Which of the following is not chiral :-A molecule is chiral if it has a carbon atom attached to four different groups (and no plane of symmetry).
3-Bromopentane: $CH_3CH_2-CH(Br)-CH_2CH_3$. Carbon 3 carries $H$, $Br$ and two identical ethyl groups, so it is not a chiral carbon, and the molecule has a plane of symmetry.
So 3-bromopentane is not chiral.
2-Butanol: $CH_3-\overset{*}{C}H(OH)-CH_2CH_3$. Carbon 2 has $H$, $OH$, $CH_3$ and $C_2H_5$, so it is chiral.
2,3-Dibromopentane: $CH_3-\overset{*}{C}HBr-\overset{*}{C}HBr-CH_2CH_3$ has two different chiral carbons and no plane of symmetry, so it is chiral.
2-Hydroxypropanoic acid (lactic acid): $CH_3-\overset{*}{C}H(OH)-COOH$ has $H$, $OH$, $CH_3$ and $COOH$ on carbon 2, so it is chiral.
Hence the compound that is not chiral is 3-bromopentane. -
1.00 g of a non-electrolyte solute (molar mass 250 g mol$^{-1}$) was dissolved in 51.2 g of benzene. If the freezing point depression constant $K_f$ of benzene is 5.12 K kg mol$^{-1}$, the freezing point of benzene will be lowered by :-$\Delta T_f = K_f \times$ molality
Molality $= \dfrac{w_{solute} \times 1000}{M_{solute} \times w_{solvent}}$
$= \dfrac{1 \times 1000}{250 \times 51.2}$ mol kg$^{-1}$
$\Delta T_f = \left(\dfrac{1 \times 1000}{250 \times 51.2}\right) \times 5.12$
$\Delta T_f = \dfrac{5120}{12800}$
$\Delta T_f = 0.4$ K -
Which of the following pairs constitutes a buffer :-A buffer solution resists a change in pH on the addition of a small amount of acid or base.
An acidic buffer is a mixture of a weak acid and its salt with a strong base; a basic buffer is a mixture of a weak base and its salt with a strong acid.
$HNO_2$ is a weak acid and $NaNO_2$ is its salt with the strong base NaOH, so this pair constitutes an acidic buffer.
Added $H^+$ is removed by $NO_2^-$ ($NO_2^- + H^+ \rightarrow HNO_2$), and added $OH^-$ is removed by $HNO_2$ ($HNO_2 + OH^- \rightarrow NO_2^- + H_2O$).
NaOH is a strong base, and $HNO_3$ and HCl are strong acids; a strong acid or strong base with its salt does not form a buffer.
So the buffer is $HNO_2$ and $NaNO_2$. -
A solution of acetone in ethanol :-In pure ethanol the molecules are held together by strong intermolecular hydrogen bonds.
When acetone is added, its molecules get in between the ethanol molecules and break some of these hydrogen bonds.
The new acetone-ethanol interactions are weaker than the ethanol-ethanol interactions they replace.
Because the intermolecular attractions are weakened, the molecules escape more easily into the vapour phase.
The vapour pressure of the solution is therefore higher than that predicted by Raoult's law, with $\Delta H_{mix} \gt 0$ and $\Delta V_{mix} \gt 0$.
So a solution of acetone in ethanol shows a positive deviation from Raoult's law. -
A hypothetical electrochemical cell is shown below:
$\overset{\ominus}{A}\,|\,A^+(xM)\,||\,B^+(yM)\,|\,\overset{\oplus}{B}$
The emf measured is +0.20 V. The cell reaction is :In the cell notation the anode (negative electrode) is written on the left and the cathode (positive electrode) on the right.
$E_{cell} = +0.20$ V is positive, so $\Delta G = -nFE_{cell}$ is negative and the cell reaction as written is spontaneous.
At the anode (left, oxidation): $A \rightarrow A^+ + e^-$
At the cathode (right, reduction): $B^+ + e^- \rightarrow B$
Adding the two half reactions gives the cell reaction:
$A + B^+ \rightarrow A^+ + B$ -
Which of the following is more basic than aniline :-The basic strength of an amine depends on the availability of the lone pair on nitrogen for accepting a proton.
In aniline, $C_6H_5NH_2$, the lone pair on nitrogen is delocalised into the benzene ring by resonance ($+M$ effect), so it is less available.
In benzylamine, $C_6H_5CH_2NH_2$, the nitrogen is attached to a $CH_2$ group and not directly to the ring, so its lone pair is localised and cannot be delocalised into the ring.
Benzylamine therefore behaves like an aliphatic primary amine and is much more basic than aniline.
In diphenylamine and triphenylamine the lone pair is delocalised over two and three benzene rings respectively, so they are weaker bases than aniline.
In p-nitroaniline the electron withdrawing $-NO_2$ group pulls the lone pair further into the ring, so it is also weaker than aniline.
So the compound more basic than aniline is benzylamine. -
The correct order of the mobility of the alkali metal ions in aqueous solution is :The extent of hydration of an ion is proportional to its charge density.
$Li^+$ is the smallest alkali metal ion, so it has the highest charge density and is the most heavily hydrated.
Down the group the ionic size increases and the degree of hydration decreases: $Li^+ \gt Na^+ \gt K^+ \gt Rb^+$
Therefore the hydrated size is largest for $Li^+$ and smallest for $Rb^+$.
A bigger hydrated ion moves more slowly through the solution, so $Li^+(aq)$ has the lowest mobility.
Order of mobility: $Rb^+ \gt K^+ \gt Na^+ \gt Li^+$ -
The correct order regarding the electronegativity of hybrid orbitals of carbon is :-The electronegativity of a hybrid orbital depends on its percentage of s-character.
An s orbital is closer to the nucleus than a p orbital, so electrons in an orbital with more s-character are held more strongly by the nucleus.
s-character: $sp$ = 50 percent, $sp^2$ = 33.3 percent, $sp^3$ = 25 percent.
Greater the s-character, greater is the electronegativity of the hybrid orbital.
So the order of electronegativity is $sp \gt sp^2 \gt sp^3$.
This explains why the hydrogen of ethyne (sp carbon) is more acidic than that of ethene or ethane. -
Which of the following species has a linear shape :-$NO_2^+$ (nitronium ion): the nitrogen has two double bonds and no lone pair, $O=\overset{+}{N}=O$. It is $sp$ hybridised, hence linear, with a bond angle of $180^\circ$. It is isoelectronic with $CO_2$.
$NO_2^-$: the nitrogen is $sp^2$ hybridised with one lone pair, so the ion is bent (about $115^\circ$).
$SO_2$: the sulphur is $sp^2$ hybridised with one lone pair, so the molecule is bent (about $119^\circ$).
$O_3$: the central oxygen is $sp^2$ hybridised with one lone pair, so the molecule is bent (about $117^\circ$).
So the species with a linear shape is $NO_2^+$.