The absolute enthalphy of neutralisation of the reaction M g O ( s ) + 2 H C l ( a q ) → M g C l 2 ( a q ) + H 2 O ( l ) will be [CBSE PMT 2005]

Greater than –57.33 kJ mol –1

Chemistry - Thermodynamics - Quiz-5

The entropy values (in JK^-1 mol^-1) of H_2(g) = 130.6, Cl_2(g) = 223.0 and HCl_(g) = 186.7 at 298 K and 1 atm pressure. Then entropy change for the reaction $H_{2 (g)} + {Cl}_{2 (g)} \to 2 {HCl}_{(g)}$ is

0%

+ 540.3
0%

+ 727.3
0%

– 166.9
0%

+ 19.8

The ΔS for the vaporization of 1 mol of water is 88.3 J/mole K. The value of ΔS for the condensation of 1 mol of vapour will be

0%

88.3 J/mol K
0%

(88.3)² J/mol K
0%

– 88.3 J/mol K
0%

$\frac{1}{88 .3}$ J/mol K

The enthalpy and entropy change for a chemical reaction are

–2.5 × 10³ cal and 7.4 cal deg^–1 respectively. Predict the reaction at 298 K is

0%

Spontaneous
0%

Reversible
0%

Irreversible
0%

Non-spontaneous

Which of the following is true for the reaction $H_{2} O (l) ⇌_{}^{} H_{2} O (g)$ at 100°C and 1 atmosphere

0%

ΔE = 0
0%

ΔH = 0
0%

ΔH = ΔE
0%

ΔH = TΔS

The enthalpy of vapourization water is 386 kJ. What is the entropy of water

0%

0.5 kJ
0%

1.03 kJ
0%

1.5 kJ
0%

22.05 kJ

Identify the correct statement regarding entropy [CBSE PMT 1998; BHU 2001]

0%

At 0°C, the entropy of a perfectly crystalline substance is taken to be zero
0%

At absolute zero temperature, the entropy of a perfectly crystalline substance is +ve
0%

At absolute zero temperature, the entropy of all crystalline substances is taken to be zero
0%

At absolute zero temperature, the entropy of a perfectly crystalline substance is taken to be zero

An engine operating between 150°C and 25°C takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is [MH CET 1999]

0%

147.7 J
0%

157.75 J
0%

165.85 J
0%

169.95 J

The standard entropies of CO₂(g), C(s) and O₂(g) are 213.5, 5.690 and 205 JK^–1 respectively. The standard entropy of formation of CO₂(g) is

0%

(1) 86 JK^–1
0%

(2) 1.96 JK^–1
0%

(3) 2.81 JK^–1
0%

(4) 2.86 JK^–1

Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed. The ratio of specific heats of the mixture (C_p/C_v) will be [AFMC 2002]

0%

1
0%

2
0%

1.67
0%

1.5

The unit of entropy is -

0%

J mol^–1
0%

JK mol^–1
0%

J mol^–1 K^–1
0%

J^–1 K^–1 mol^–1

The entropy changed involved in the conversion of 1 mole of liquid water at 373 K to vapour at the same temperature will be $[Δ H_{vap} = 2 .257 k J / g m]$ [MP PET 2002]

0%

0.119 kJ
0%

0.109 kJ
0%

0.129 kJ
0%

0.120 kJ

When a liquid boils, there is [JIPMER 2002]

0%

An increase in entropy
0%

A decrease in entropy
0%

An increase in heat of vaporization
0%

An increase in free energy

The work done to contract a gas in a cylinder, is 462 joules. 128 joule energy is evolved in the process. What will be the internal energy change in the process [MP PMT 2003]

0%

+ 590 joules
0%

– 334 joules
0%

+ 334 joules
0%

– 590 joules

For a carnot engine, the source is at 500 K and the sink at 300 K. What is efficiency of this engine

0%

0.2
0%

0.4
0%

0.6
0%

0.3

From Kirchhoff's equation which factor affects the heat of reaction [MP PMT 1990]

0%

Pressure
0%

Temperature
0%

Volume
0%

Molecularity

The absolute enthalphy of neutralisation of the reaction $M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)$ will be [CBSE PMT 2005]

0%

Less than –57.33 kJ mol^–1
0%

–57.33 kJ mol^–1
0%

Greater than –57.33 kJ mol^–1
0%

57.33 kJ mol^–1

The heat of transition (ΔH_t) of graphite into diamond would be, where

$C (graphite) + O_{2} (g) \to C O_{2} (g); Δ H = x k J$

$C (diamond) + O_{2} (g) \to C O_{2} (g); Δ H = y k J$ [Pb. PET 1985]

0%

$(x + y) k J m o l^{- 1}$
0%

$(x - y) k J m o l^{- 1}$
0%

$(y - x) k J m o l^{- 1}$
0%

None of these

Given that

$2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g); Δ H = - 787 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286 k J$

$C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l); Δ H = - 1301 k J$

Heat of formation of acetylene is

0%

–1802 kJ
0%

+1802 kJ
0%

–800 kJ
0%

+228 kJ

The enthalpy of combustion at 25°C of H₂, cyclohexane (C₆H₁₂) and cyclohexene (C₆H₁₀) are –241, –3920 and –3800 KJ / mole respectively. The heat of hydrogenation of cyclohexene is

0%

– 121 KJ / mole
0%

+ 121 KJ / mole
0%

– 242 KJ / mole
0%

+ 242 KJ / mole

The heat change for the reaction $H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O$ (l) is called

0%

Heat of reaction
0%

Heat of formation
0%

Heat of vaporization
0%

None of the above

The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol^–1. The heat released when 0.5 mole of HNO₃ solution is mixed with 0.2 mole of KOH is [KCET 1991; AIIMS 2002;

0%

57.0 kJ
0%

11.4 kJ
0%

28.5 kJ
0%

34.9 kJ

The enthalpy of neutralization of HCN by NaOH is –12.13 kJ mol^–1. The enthalpy of ionisation of HCN will be -

0%

45.07 kJ
0%

4.310 kJ
0%

451.9 kJ
0%

33.77 kJ

A solution of 500 ml of 0.2 M KOH and 500 ml of 0.2 M HCl is mixed and stirred; the rise in temperature is T₁. The experiment is repeated using 250 ml each of solution, the temperature raised is T₂. Which of the following is true

0%

T₁ = T₂
0%

T₁ = 2T₂
0%

T₁ = 4T₂
0%

T₂ = 9T₁

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form, ΔH is –453.5 cal. This points out that

0%

Graphite is chemically different from diamond
0%

Graphite is as stable as diamond
0%

Graphite is more stable than diamond
0%

Diamond is more stable than graphite

Which of the following equations correctly represents the standard heat of formation $(Δ H_{f}^{o})$ of methane [IIT JEE (Screening) 1992]

0%

$C (d i a m o n d) + 2 H_{2} (g) = C H_{4} (g)$
0%

$C (g r a p h i t e) + 2 H_{2} (g) = C H_{4} (l)$
0%

$C (g r a p h i t e) + 2 H_{2} (g) = C H_{4} (g)$
0%

$C (g r a p h i t e) + 4 H = C H_{4} (g)$

In which of the following reactions does the heat change represent the heat of the formation of water

0%

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); Δ H = - 116 k c a l$
0%

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 58 k c a l$
0%

$H^{+} (a q .) + O H^{-} (a q .) \to 2 H_{2} O (l); Δ H = - 13 .7 k c a l$
0%

$C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)$ ; $Δ H = - 310 k c a l$

Based on the following thermochemical equations

$H_{2} O (g) + C (s) \to C O (g) + H_{2} (g); Δ H = 131 k J$

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H = - 282 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 242 k J$

$C (s) + O_{2} (g) \to C O_{2} (g); Δ H = X k J$

The value of X is [CBSE PMT 1992]

0%

–393 kJ
0%

–655 kJ
0%

+393 kJ
0%

+655 kJ

If enthalpies of formation of C₂H₄(g), CO₂(g) and H₂O(l) at 25°C and 1 atm pressure be 52, – 394 and –286 kJ mol^–1 respectively, the enthalpy of combustion of C₂H₄(g) will be [CBSE PMT 1995; AIIMS 1998; Pb. PMT 1999]

0%

+1412 kJ mol^–1
0%

–1412 kJ mol^–1
0%

+141.2 kJ mol^–1
0%

–141.2 kJ mol^–1

Ozone is prepared by passing silent electric discharge through oxygen. In this reaction [AFMC 1998]

0%

Energy is given out
0%

Energy is absorbed
0%

Oxygen is loaded with energy
0%

Oxygen is dissociated into atoms

Combustion of glucose takes place according to the equation,

$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 {CO}_{2} + 6 H_{2} O$ , $∆ H = - 72 K c a l$

Energy required for the production of 1.6 g of glucose is -

(Molecular mass of glucose = 180 g)

0%

0.064 kcal
0%

0.64 kcal
0%

6.4 kcal
0%

64 kcal

0:0:1

Practice Chemistry Quiz Questions and Answers

0:0:1

Practice Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.