The absolute enthalpy of neutralisation of the reaction
MgO(s) + 2HCl(aq) $\to $ MgCl_{2}(aq) + H_{2}O(l)
will be
less than - 57.33 kJ mol^{-1}
- 57.33 kJ mol^{-1}
greater than-- 57.33 kJ mol^{-1}
57.33 kJ mol^{-1}
A.
less than - 57.33 kJ mol^{-1}
Heat of neutralisation of strong acid and strong base is -57.33 kJ. MgO is weak base while HCl is strong acid, so the heat of neutralisation of MgO and HCl is lower to -57.33 kJ because MgO requires some heat in ionisation, then net released amount of heat is decreased.
A reaction occurs spontaneously if
T$\u2206$S < $\u2206$H and both $\u2206$H and $\u2206$S are +ve
T$\u2206$S > $\u2206$H and both $\u2206$H and $\u2206$S are +ve
T$\u2206$S = $\u2206$H and both $\u2206$H and $\u2206$S are +ve
T$\u2206$S > $\u2206$H and $\u2206$H is +ve and $\u2206$S is - ve
B.
T$\u2206$S > $\u2206$H and both $\u2206$H and $\u2206$S are +ve
The spontaneity of reaction is based upon the negative value of $\u2206$G, $\u2206$G is based upon T, $\u2206$S and $\u2206$H according to the following equation (Gibbs-Helmholtz equation):
$\u2206\mathrm{G}=\u2206\mathrm{H}-\mathrm{T}\u2206\mathrm{S}$
If the magnitude of $\u2206$H - T$\u2206$S is negative, then the reaction is spontaneous.
When T$\u2206$S > $\u2206$H and $\u2206$H and $\u2206$S are +ve, then $\u2206$G is negative.
The number of moles of KMnO_{4} reduced by one mole of KI in alkaline medium is
one-fifth
five
one
two
D.
two
In alkaline solution, KMnO_{4} is reduced to MnO_{2} (colourless).
$2{\mathrm{KMnO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}\to 2{\mathrm{MnO}}_{2}+2\mathrm{KOH}\phantom{\rule{0ex}{0ex}}\overline{)+3\left[\mathrm{O}\right]\mathrm{Ki}+3\left[\mathrm{O}\right]\to {\mathrm{KIO}}_{3}}\phantom{\rule{0ex}{0ex}}2{\mathrm{KMnO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}+\mathrm{KI}\to 2{\mathrm{MnO}}_{2}+2\mathrm{KOH}+{\mathrm{KIO}}_{3}$
Hence, two moles of KMnO_{4} are reduced by one mole of KI.
Which of the following molecules has trigonal planar geometry?
IF_{3}
PCl_{3}
NH_{3}
BF_{3}
D.
BF_{3}
Among all the given options, IF_{3} has trigonal planar geometry.
PCl_{3} has pyramidal geometry.
NH_{3} has pyramidal geometry.
BF_{3} has trigonal planar geometry
Which of the following is the electron deficient molecule?
B_{2}H_{6}
C_{2}H_{6}
PH_{3}
SiH_{4}
A.
B_{2}H_{6}
B_{2}H_{6} is electron deficient molecule because boron atom has three half filled orbitals in excited state. The structure of B_{2}H_{6} is represented as follows :
In it two electrons of a B - H bond are involved in formation of three centre bond, these bonds are represented as dotted lines.
Equilibrium constants K_{1} and K_{2} for the following equilibria
NO (g) + $\frac{1}{2}{\mathrm{O}}_{2}$ $\stackrel{{\mathrm{K}}_{1}}{\rightleftharpoons}$ NO_{2}(g) and
2NO_{2}(g) $\stackrel{{\mathrm{K}}_{2}}{\rightleftharpoons}$ 2NO (g) + O_{2}(g)
are related as
${\mathrm{K}}_{2}=\frac{1}{{\mathrm{K}}_{1}}$
${\mathrm{K}}_{2}={\mathrm{K}}_{1}^{2}$
${\mathrm{K}}_{2}=\frac{{\mathrm{K}}_{1}}{2}$
${\mathrm{K}}_{2}=\frac{1}{{\mathrm{K}}_{1}^{2}}$
D.
${\mathrm{K}}_{2}=\frac{1}{{\mathrm{K}}_{1}^{2}}$
(i) NO (g) + $\frac{1}{2}{\mathrm{O}}_{2}$(g) $\stackrel{{\mathrm{K}}_{1}}{\rightleftharpoons}$ NO_{2}(g)
So, K_{1} = $\frac{\left[{\mathrm{NO}}_{2}\right]}{\left[\mathrm{NO}\right][{\mathrm{O}}_{2}{]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ ...(1)
(ii) 2NO_{2} (g) $\stackrel{{\mathrm{K}}_{2}}{\rightleftharpoons}$ 2NO + O_{2} (g)
So, K_{2} = $\frac{[\mathrm{NO}{]}^{2}[{\mathrm{O}}_{2}]}{\left[{\mathrm{NO}}_{2}\right]}$ ...(2)
From equation (1)
${\mathrm{K}}_{1}^{2}=\frac{[{\mathrm{NO}}_{2}{]}^{2}}{\left[{\mathrm{NO}}_{2}\right]}$ or
$\frac{1}{{\mathrm{K}}_{1}^{2}}=\frac{\left[\mathrm{NO}{]}^{2}\right[{\mathrm{O}}_{2}]}{[{\mathrm{NO}}_{2}{]}^{2}}$ ...(3)
from equation (2) and (3)
K_{2} = $\frac{1}{{\mathrm{K}}_{1}^{2}}$
The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (Atomic mass Al= 27)
180 kg
270 kg
540 kg
90 kg
D.
90 kg
In Hall and Heroult process
2Al_{2}O_{3} $\to $ 4Al + 3O_{2}
4C + 3O_{2} $\to $ 2CO_{2} + 2CO $\uparrow $
2Al_{2}O_{3} + 4C $\to $4Al + 2CO_{2} + 2CO
Only for removal of CO_{2}, following equation is possible
2Al_{2}O_{3} + 3C $\to $4Al + 3CO_{2}
Molecular weight of Carbon = 12
Molecular weight of Aluminium = 27
$\therefore $ For Carbon/ C = 3 x 12 = 36
For Aluminium/ Al = 4 x 27 = 108
$\because $ For 108 gm of Al, 36 gm of C is required in above reaction
$\therefore $ For 270 gm of Al require amount of C = $\frac{36}{108}\times 270=90\mathrm{gm}$
A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three $\mathrm{\alpha}$-particles in succession. The group of the periodic table to which the resulting daughter element would belong to
Group 14
Group 16
Group 4
Group 6
A.
Group 14
After decay of radioactive element, the last element Pb is obtained which belongs to group 14 of periodic table.
Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?
Exothermic and decreasing disorder
Endothermic and increasing disorder
Exothermic and increasing disorder
Endothermic and decreasing disorder
C.
Exothermic and increasing disorder
If a reaction is exothermic, $\u2206$H will be -ve and in increasing order, $\u2206$S will be +ve thus, at this condition, $\u2206$G is negative according to following equation:
$\u2206$G = $\u2206$H - T$\u2206$S
$\u2206$G = -ve.
Hence, for spontaneous reaction $\u2206$G must be -ve
The correct sequence of increasing covalent character is represented by
LiCl < NaCl < BeCl_{2}
BeCl_{2} < NaCl < LiCl
NaCl < LiCl < BeCl_{2}
BeCl_{2} < LiCl < NaCl
C.
NaCl < LiCl < BeCl_{2}
According to Fajan's rule, lower the size of cation higher will be its polarising power and higher will be covalent character.
$\therefore $ Polarising power $\propto $ $\frac{1}{\mathrm{size}\mathrm{of}\mathrm{action}}$
Covalent character $\propto $Polarising power
So, the correct order is NaCl < LiCl < BeCl_{2}
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