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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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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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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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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 the P-V diagram given for an ideal gas, out of the following which one correctly represents the T-P diagram ?  
Option: 1

Option: 2 

Option: 3 

Option: 4 
 

\because PV=constant

\therefore The process is isothermal (temperature constant)

Also, from the process given, it can be seen that pressure is decreasing, P2<P1

Hence, the correct answer is Option (3)

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vishal kumar

A steel rail of length 5 m and area of cross section 40 cm2 is prevented from expanding along its length while the temperature rises by 100C.  If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately :
Option: 1 2×107 N
Option: 2  1×105 N
Option: 3  2×109 N
Option: 4  3×10−5 N  
 

\\ \text{Youngs modules} \ \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2} \\ \text{linears expression coefficient} \alpha=1.2 \times 10^{-5} \mathrm{~K}^{-1} \\ 1=5 \mathrm{~cm} \quad \mathrm{~A}=40 \mathrm{~cm}^2 \\ \Delta \mathrm{T}=10^{\circ} \mathrm{C} \\ \Delta \mathrm{l}=\mathrm{l} \propto \Delta \mathrm{T} \\ \frac{\Delta \mathrm{l}}{\mathrm{l}}=\propto \Delta \mathrm{T} \left[\mathrm{F}=\mathrm{YA} \frac{\Delta \mathrm{l}}{\mathrm{l}}\right] \\ \mathrm{F}=\mathrm{YA} \propto \Delta \mathrm{T} \mathrm{F}=2 \times 10^{11} \mathrm{Nm}^{-2} \times \frac{40}{10} \\mathrm{~m} \times 1.25 \times 10^{-5} \times 10 =2 \times 40 \times 12 \times 10^2 =96 \times 10^3 =0.96 \times 10^3 F \cong 1 \times 10^5 \mathrm{~N}

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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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vishal kumar

 In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 1500C. Immediately, it is put into water of volume 150 cc at 270C kept in a calorimeter of water equivalent to 0.025 kg.  Final temperature of the system is 400C.  The specific heat of aluminium is : (take 4.2 Joule=1 calorie)
Option: 1 378 J/kg – 0C
Option: 2  315 J/kg – 0C
Option: 3  476 J/kg – 0C  
Option: 4  434 J/kg – 0C  
 

\\ \text{Heat lost by metal Blodg} = \text{Heat gain by} \mathrm{H}_2 \mathrm{O}+ \text{contain} \\ \begin{aligned} &\Rightarrow \mathrm{m}_1 \mathrm{C}_1 \Delta \mathrm{T}_1=\left(\mathrm{m}_2+\mathrm{w}\right) \mathrm{C}_2 \Delta \mathrm{T}_2 \\ &\Rightarrow 0.20 \times \mathrm{C}_1 \times(150-40)=\left(150 \times 10^{-3}\right) \\ &\mathrm{C}_1=434 \mathrm{y} \mathrm{kg}^{-1} \mathrm{k}^{-1} \end{aligned}

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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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vishal kumar

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