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#### The heats of combustion of carbon and carbon monoxide are −393.5 and −283.5 kJ mol−1, respectively.  The heat of formation (in kJ) of carbon monoxide per mole is: Option: 1 110.5 Option: 2 676.5 Option: 3 -676.5 Option: 4 -110

$\mathrm{C_s+O_2_g \rightarrow CO_2_g \: \: \: \: \Delta H= -393.5 KJmol^{-1}}$

$CO_{\left ( g \right )}+\frac{1}{2}O_{2} \: _{\left ( g \right )}\rightarrow CO_{2} \:_{(g)} \: \: \: \: \: \Delta H=-283.5 kJmol^{-1}$

$C_{\left ( s \right )}+\frac{1}{2}O_{2} \:_{\left ( g \right )}\rightarrow CO_{\left ( g \right )}$

$\therefore \Delta H= -393.5+283.5$

$= -110.0\ kJmol^{-1}$

Therefore, Option(4) is correct

#### The group having triangular planar structures is :   Option: 1 Option: 2  Option: 3    Option: 4

The group having triangular planar structures is CO32-, NO?????3?-,SO?3
In CO32- ion, C atom is sp?????2 hybridised. This results in triangular planar structure.
In NO?????3????- ion, N atom is sp?????2 hybridised. This results in triangular planar structure.
In SO?????3, S atom is sp?????2 hybridised. This results in triangular planar structure.
In all above molecules/ions, the central atom has 3 bonding domains and bond angle of 120o.

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#### A gas undergoes change from state A to state B.  In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively.  Now gas is brought back to A by another process during which 3 J of heat is evolved.  In this reverse process of B to A : Option: 1  10 J of the work will be done by the gas. Option: 2  6 J of the work will be done by the gas. Option: 3 10 J of the work will be done by the surrounding on gas. Option: 4  6 J of the work will be done by the surrounding on gas.

$A \longrightarrow B$

$\mathrm{Q=5 \ J}$

$\mathrm{W=8\ J}$

$\mathrm{\Delta U_{AB}=Q+W=5+(-8)=-3J}$

$B \longrightarrow A$

$\mathrm{Q=-3J}$

$\mathrm{\Delta U_{BA}=3J}$

$\Delta U_{BA}=-3+W$

$3+3=W$

$W=6J$

(work is done on the system)

Or

As work done has a positive sign, work is done by the surrounding on the gas.

Hence, the correct answer is (4)

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#### U is equal to : Option: 1 Adiabatic work Option: 2  Isothermal work Option: 3  Isochoric work Option: 4 Isobaric work

Heat exchange between the system and surroundings is zero.

So,

$\Delta E= q+w$

$q= 0$

$\Delta E=w$

No change in internal energy = Adiabatic work

Ans(1)

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#### Given C(graphite)+O2(g) → CO2(g) ; rH0=−393.5 kJ mol−1 H2(g)+ O2(g) → H2O(l) ; rH0=−285.8 kJ mol−1 CO2(g)+2H2O(l) → CH4(g)+2O2(g) ; rH0=+890.3 kJ mol−1 Based on the above thermochemical equations, the value of rH0 at 298 K for the reaction C(graphite)+2H2(g) → CH4(g) will be : Option: 1  −74.8 kJ mol−1 Option: 2 −144.0 kJ mol−1 Option: 3 +74.8 kJ mol−1 Option: 4 +144.0 kJ mol−1

$C+O_{2} \rightarrow CO_{2} , \Delta H = -3.93.5$ kJ mol−1

$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O , \Delta H =28.5$ kJ mol−1

$\Delta 2H_{2}+O_{2}\rightarrow 2H_{2}O, \Delta H= 2\times \left ( -285.8 \right )$ kJ mol−1

$CO_{2}+2H_{2}O \rightarrow CH_{4}+2O_{2}, \Delta H=890.3$ kJ mol−1

$C+2H_{2}\rightarrow CH_{4}$

$\Delta H = 890.3-2\times 285.8-399.6$

$= -74.8$ kJ mol−1

#### Among the following, the set of parameters that represents path functions,is :$\left ( A \right )$  $q+w$ $\left ( B \right )$   $q$ $\left ( C \right )$   $w$ $\left ( D \right )$  $H-TS$ Option: 1    Option: 2    Option: 3   Option: 4

$Q+w=\Delta U$   is state function

$H-TS=G$  is state function

q and w are path functions

because q and w depend on the path of the reaction.

Hence, option number (1) is correct.

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#### During the nuclear explosion, one of the products is $^{90}Sr$  with a half-life of 6.93 years. If was absorbed in the bones of a newly born baby in place of Ca, how much time, in the year, is required to reduce it by 90% if it is not lost metabolically----------- Option: 1 23.03 Option: 2 46.06 Option: 369.06 Option: 422.66

We know this formula:-

$\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[R]}}$

For 90% completion, the required time will be -

$\\\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[0.1R_{o}]}}$

$\\\mathrm{t\: =\: \frac{2.303}{k}log10}$

$\\\mathrm{t\: =\: \frac{2.303}{k}}$

Half-life is given as 6.93 years. So,
$\\\mathrm{k\: =\: \frac{0.693}{t_{\frac{1}{2}}}}$

$\\\mathrm{k\: =\: \frac{0.693}{6.93}\: =\: 0.1}$

Thus, t is given as fro 90% completion:

$\\\mathrm{t\: =\: \frac{2.303}{k}}$

$\\\mathrm{t\: =\: \frac{2.303}{0.1}}$

Thus, t = 23.03 years

#### For the reaction $\\A(l)\rightarrow 2B(g)\\\Delta U=2.1\: kcal,\Delta S=20\: cal\: K^{-1}\: at\; 300\; K\\Hence\; \Delta G\: in\: kcl\: is-----$ Option: 1 -2.7 Kcal Option: 2 3.3 Kcal Option: 3 - 3.3 Kcal Option: 42.7 Kcal

We know:

$\mathrm{\Delta G\: =\: \Delta H\: -\: T\Delta S}$

$\mathrm{\Delta H\: =\: \Delta U\: +\: 2RT}$

Thus, we have:

$\mathrm{\Delta G\: =\: \Delta U\: +\: 2RT\: -\: T\Delta S}$

On putting the given values we get:

$\mathrm{\Delta G\: =\: 2.1\: +\: 2\, x\, 2\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\: 2.1\: +\:4\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\: 1200\, x\, 10^{-3}\: -\: 6000\, x\, 10^{-3}}$

$\mathrm{\Delta G\: =\:2.1\: +\: 1.2\: -\: 6}$

$\mathrm{\Delta G\: =\:3.3\: -\: 6}$

$\mathrm{\Delta G\: =\:-2.7\ Kcal}$

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