When 0.16 g of glucose was burnt in a bomb calorimeter, the temperature rose by 4 deg. Calculate the calorimeter constant (water equivalent of the calorimeter) given that $∆H^{\circ }=-2.8\times {10}^6<mspace\; width="1em"></mspace>J<mspace\; width="1em"></mspace>mo{l}^{-1}.$ [molar enthalpy of combustion]. Molar mass of glucose = 180 mol^-1.

1. $5.73\times {10}^2<mspace\; width="1em"></mspace>J/deg$

2. $7.53\times {10}^2<mspace\; width="1em"></mspace>J/deg$

3. $6.22\times {10}^2<mspace\; width="1em"></mspace>J/deg$

4. $3.57\times {10}^2<mspace\; width="1em"></mspace>J/deg$

The C-Cl bond energy can be calculated from :

1. $∆{H^{\circ }}_f\left(CC{l}_4,<mspace\; width="1em"></mspace>l\right)<mspace\; width="1em"></mspace>only$

2. ${{∆}^{\circ }}_f\left(CC{l}_4,<mspace\; width="1em"></mspace>l\right)<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>∆\left(C{l}_2\right)$

3. $∆{H^{\circ }}_f\left(CC{l}_4,<mspace\; width="1em"></mspace>l\right)<mspace\; width="1em"></mspace>∆\left(C{l}_2\right)$

4. $∆{H^{\circ }}_f\left(CC{l}_4,<mspace\; width="1em"></mspace>l\right)<mspace\; width="1em"></mspace>∆\left(C{l}_2\right),<mspace\; width="1em"></mspace>∆{H^{\circ }}_f\left(C,<mspace\; width="1em"></mspace>g\right)<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>∆{H^{\circ }}_{vap}\left(CC{l}_4\right)$

Given $∆{H^{\circ }}_f$ of DyCl₃ (s) = -994.30 kJ mol^-1

$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\stackrel{+aq}{\to }HCl\left(aq.<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>M\right);<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>∆H^{\circ }=-158.31<mspace\; width="1em"></mspace>kJ<mspace\; width="1em"></mspace>mo{l}^{-1}$
$DyC{l}_3\left(s\right)\underset{HCl}{\overset{aq}{\to }}DyC{l}_3\left(aq.<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>HCl\right);<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>∆H^{\circ }=-180.06<mspace\; width="1em"></mspace>kJ<mspace\; width="1em"></mspace>mo{l}^{-1}$
$Dy\left(s\right)\underset{aq.<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>M}{\overset{+3<mspace\; width="1em"></mspace>HCl}{\to }}DyC{l}_3\left(aq.<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>M<mspace\; width="1em"></mspace>HCl\right)+\frac{3}{2}{H}_2\left(g\right);<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>∆H^{\circ }=x,<mspace\; width="1em"></mspace>calculate<mspace\; width="1em"></mspace>x.$
$$

1. -966.5 kJ/mol

2. -699.43 kJ/mol

3. -596.6 kJ/mol

4. -569.6 kJ/mol

When 1 g H₂ gas at S.T.P is expanded to twice its initial volume, then the work done is:

1. 22.4 L atm
2. 5.6 L atm
3. 11.2 L atm
4. 44.8 L atm

The gas absorbs 100 J heat and is simultaneously compressed by a constant external pressure of 1.50 atm from 8 lit. to 2 lit. in volume. Hence $∆E$ will be-

1. -812 J

2. 812 J

3. 1011 J

4. 911 J

If $∆G=∆H-T∆S$ and $∆G=∆H+T{\left(\frac{d\left(∆G\right)}{dT}\right)}_P$ then variation of EMF of a cell E, with temperature T, is given by

1. $\frac{∆H}{nF}$ 

2. $\frac{∆G}{nF}$

3. $\frac{∆S}{nF}$

4. $-\frac{∆S}{nF}$

The standard heat of combustion of Al is -837.8 kJ mol^-1 at ${25}^{\circ }C$ which of the following releases 250 kcal of heat ?

1. The reaction of 0.624 mol of Al

2. The formation of 0.624 mol of Al₂O₃

3. The reaction of 0.312 mol of Al

4. The formation of 0.150 mol of Al₂O₃

$C_P-C_V=R$. This R is :

1. Change in K.E.

2. Change in rotational energy

3. work done which system can do on expanding the gas per mol per degree increase in temperature

4. All correct

Heat of neutralisation of oxalic acid is -25.4 K cal mol^-1 using strong base, NaOH. Hence enthalpy change of the process is $H_2{C}_2{O}_4\rightleftharpoons 2H^{+}+C_2{H_4}^{2-}<mspace\; width="1em"></mspace>$is-

1. 2.0 kcal

2. -11.8 kcal

3. 1.0 kcal

4. -1.0 kcal

5 moles of nitrogen gas are enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston.

The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly :

1. 2K/sec

2. 4K/sec

3. 6K/sec

4. 8K/sec