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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 $∆ H^{\circ} = - 2.8 \times 10^{6} J m o l^{- 1} .$ [molar enthalpy of combustion]. Molar mass of glucose = 180 mol^-1.

$5.73 \times 10^{2} J / d e g$

$7.53 \times 10^{2} J / d e g$

$6.22 \times 10^{2} J / d e g$

$3.57 \times 10^{2} J / d e g$

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Explanation

180 gms of glucose $\Rightarrow$ $2.8 \times 10^{6} J$ of heat evolved

$∴$ 0.16 gms would yield $\frac{2.8 \times 10^{6}}{180} \times 0.16 J$

If the calorimeter constant = W, then

$W \times 4 = \frac{2.8 \times 0.16 \times 10^{6}}{180} = \frac{2.8 \times 1.6 \times 10^{3}}{1.8} J ∴ W = \frac{2.8 \times 1.6 \times 10^{3}}{1.8 \times 4} J d e g^{- 1} = 6.22 \times 10^{2} J d e g^{- 1}$

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The C-Cl bond energy can be calculated from :

$∆ {H^{\circ}}_{f} (C C l_{4}, l) o n l y$

${∆^{\circ}}_{f} (C C l_{4}, l) a n d ∆ (C l_{2})$

$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2})$

$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2}), ∆ {H^{\circ}}_{f} (C, g) a n d ∆ {H^{\circ}}_{v a p} (C C l_{4})$

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Explanation

$C (s) + 2 C l_{2} (g) \to C C l_{4} (l) ∆ H_{f} (C C l_{4}, l) = ∆ H^{\circ} [C (s) \to C (g)] + 2 {(B E)}_{C l - C l} - [∆ {H^{\circ}}_{v a p} (C C l_{4}) + 4 {(B E)}_{C l - C l}]$

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

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \overset{+ a q}{\to} H C l (a q . 4 M); ∆ H^{\circ} = - 158.31 k J m o l^{- 1} D y C l_{3} (s) \to_{H C l}^{a q} D y C l_{3} (a q . i n 4 M H C l); ∆ H^{\circ} = - 180.06 k J m o l^{- 1} D y (s) \to_{a q . 4 M}^{+ 3 H C l} D y C l_{3} (a q . 4 M H C l) + \frac{3}{2} H_{2} (g); ∆ H^{\circ} = x, c a l c u l a t e x .$

-966.5 kJ/mol

-699.43 kJ/mol

-596.6 kJ/mol

-569.6 kJ/mol

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Explanation

$D y (s) + \frac{3}{2} C l_{2} (g) ⇌ D y C l_{3} (s) ∆ H = - 994.30 k J m o l^{- 1} D y C l_{3} (s) \overset{a q . H C l}{\to} D y C l_{3} (a q . i n 4.0 M H C l) ∆ H = - 180.06 k J m o l^{- 1} 3 H C l (a q . 4 M) \to \frac{3}{2} H_{2} (g) + \frac{3}{2} C l_{2} (g) ∆ H = 3 \times 158.31 K J m o l^{- 1} D y (s) + 3 H C l ⇌ D y C l_{3} + \frac{3}{2} H_{2} (g) ∆ H = - 699.43 k J m o l^{- 1} = x (a q . 4 M) (a q . 4 M H C l)$

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1 g H₂ gas at S.T.P is expanded so that volume is doubled. Hence work done is:

22.4 L atm

5.6 L atm

11.2 L atm

44.8 L atm

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Explanation

V₁(volume of 1 g H₂) = 11.2 L at NTP

V₂(volume of 1 g H₂) = 22.4 L

$∴ W = P ∆ V = 11.2 L a t m$

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$∆ H$ for the reaction 2C(s) + 3H₂(g) $\to$ C₂H₆(g) is -20.24 kcal/mol. To what value of the enthalpy of sublimation of C(s) does this point given that the bond energies of C-C, C-H and H-H are 63 kcal/mol, 85.6 kcal/mol and 102.6 kcal/mol.

124.3 kcal/mol

185.3 kcal/mol

158.3 kcal/mol

211.5 kcal/mol

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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-

-812 J

812 J

1011 J

911 J

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Explanation

$∆ H = ∆ E + P ∆ V 100 = ∆ E + 1.5 (2 - 8) \times \frac{8.314}{0.0821} ∆ E = 1011.4 J$

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

$\frac{∆ H}{n F}$

$\frac{∆ G}{n F}$

$\frac{∆ S}{n F}$

$- \frac{∆ S}{n F}$

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Explanation

On comparison : $∆ S = [\frac{d (∆ G)}{d T}] ∆ S = \frac{d (- n F E)}{d T} = n F (\frac{d E}{d T}) ∴ (\frac{d E}{d T}) = \frac{∆ S}{n F}$

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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 ?

The reaction of 0.624 mol of Al

The formation of 0.624 mol of Al₂O₃

The reaction of 0.312 mol of Al

The formation of 0.150 mol of Al₂O₃

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Explanation

$A l + \frac{3}{4} O_{2} = \frac{1}{2} A l_{2} O_{3}; ∆ H = - 837.8 k J 250 k c a l = 250 \times 4.2 k J = 1050 k J 1050 k J h e a t i n d i c a t e d t h e f o r m a t i o n o f \frac{1}{2} \times \frac{1050}{837.8} = 0.624 m o l e s o f A l_{2} O_{3} .$

Thus, the liberation of 250 kcal energy indicating the formation of 0.624 moles of Al₂O₃

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$C_{P} - C_{V} = R$ . This R is :

Change in K.E.

Change in rotational energy

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

All correct

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Explanation

PV=RT at temp T for one mol

$P (V + ∆ V) = R (T + 1)$ at temp. (T+1) for one mol

$∴ P ∆ V = R$

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Heat of neutralisation of oxalic acid is -25.4 K cal mol^-1 using strong base, NaOH. Hence enthalpy change of the process is $H_{2} C_{2} O_{4} ⇌ 2 H^{+} + C_{2} {H_{4}}^{2 -}$ is-

2.0 kcal

-11.8 kcal

1.0 kcal

-1.0 kcal

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Explanation

The heat of neutralisation of strong acid and strong base = -13.7 kcal/equiv.

$∴ - 25.4 = - 2 \times 13.7 \times ∆ H_{d i s s o} o r ∆ H_{d i s s o} = 2 \times 13.7 - 25.4 = 2 k c a l$

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