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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

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Posted by

Ritika Jonwal

 The group having triangular planar structures is :  
Option: 1BF_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 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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Posted by

vishal kumar

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 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

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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vishal kumar

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 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

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

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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

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

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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

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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Kuldeep Maurya

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