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No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

2

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Posted by

Mura

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

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Posted by

Ritika Jonwal

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The group having triangular planar structures is :
Option: 1 $BF_3, NF_3,CO_3^{2-}$
Option: 2 $CO ^{2-}_3, NO_3^-, SO_3$
Option: 3 $NH_3, SO_3,CO^{2-}_3$
Option: 4 $NCl_3, BCl_3, SO_3$

The group having triangular planar structures is CO₃^2-, NO???_??3^?-,SO_?3
In CO₃^2- 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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Posted by

vishal kumar

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : (Assume activation energy and preexponential factor are independent of temperature; ln 2=0.693; R=8.314 J mol⁻¹ K⁻¹)
Option: 1 107.2 kJ mol⁻¹
Option: 2 53.6 kJ mol⁻¹
Option: 3 26.8 kJ mol⁻¹
Option: 4 214.4 kJ mol⁻¹

$\\\text{The rate of the reaction can be written as:}\\ \\ ln\frac{K_{2}}{K_{1}}= \frac{E_{a}}{R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}]\\ \\ ln\frac{4}{1}= \frac{E_{a}}{8.314}[\frac{1}{300}-\frac{1}{310}]\\ \\ E_{a}= 107.2 \ kJ/mol$

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Posted by

vishal kumar

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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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Posted by

vishal kumar

For a reaction, A(g) → A( $\small \l$ ); $\mathrm{\Delta H = -3RT}$ . The correct statement for the reaction is :
Option: 1 $|\Delta \mathrm{H}|<|\Delta \mathrm{U}|$
Option: 2 $\Delta H= \Delta U= 0$
Option: 3 $|\Delta \mathrm{H}|>|\Delta \mathrm{U}|$

Option: 4 $|\Delta \mathrm{H}|=|\Delta \mathrm{U}|$

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Posted by

vishal kumar

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

Adiabatic Process -

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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Posted by

vishal kumar

Given C_(graphite)+O₂(g) → CO₂(g) ; _rH⁰=−393.5 kJ mol⁻¹ H₂(g)+ $\frac{1}{2}$ 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

Adding all equations, we get

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

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

$= -74.8$ kJ mol⁻¹

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Posted by

vishal kumar

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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    $\left ( B \right )\: and\: \left ( C \right )$
Option: 2    $\left ( B \right ),\left ( C \right )and \left ( D \right )$
Option: 3 $\left ( A \right )\: and \: \left ( D \right )$
Option: 4 $\left ( A \right ),\left ( B \right ) and \left ( C \right )$

$Q+w=\Delta U$ 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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Posted by

Ritika Jonwal

During the nuclear explosion, one of the products is $^{90}Sr$ with a half-life of 6.93 years. If $\mathrm{1\ \mu g\: of\: ^{90}Sr}$ 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

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Posted by

Kuldeep Maurya

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