Friday, 3 August 2012

11. ERQs of Thermochemistry


S.No

Question


Consider two metals A and B , each having a mass of 100 g and initial temperature of 20 0C.The specific heat of A is larger than that B .Under the same heating conditions ,   which metal would take longer to reach a temperature  of 21 0C

A quantity of 2.00 x102   ml of 0.862M HCl is mixed with 2.00 x 10ml of 0.431M Ba(OH)2 in a constant- pressure Calorimeter  that has a heat capacity is 453J/0C .The initial temperature of HCl and  Ba(OH)2  solution is the same at 20.48 0C, the of neutralization is -56.2 kJ. What is the final temperature of the solution.

From the following heats of combustion
CH3OH (l) + 3/2  O2g) → CO2 (g) 
                                                ∆H=-726kJ
C (graphite) + O2(g) → CO2 (s)     ∆H=-393kJ
H2(g) + ½  O2(g) → H2O (l)     ∆H=-285.8kJ

Calculate the enthalpy of formation of methanol (CH3OH)
C (graphite) + ½ O2(g)  +2 H2(g) →CH3OH(l)

From the following heats of combustion
C2H2 (g) + 7 O2g) →4 CO2 (g) + 6H2O(l)                                                                 
                                             ∆H=-3119.6kJ
C (graphite) + O2(g) → CO2 (s)   ∆H=-393kJ
H2(g) + ½  O2(g) → H2O (l)     ∆H=-285.8kJ

Calculate the enthalpy of formation of C2H2 (g)
 2C (graphite) + 2 H2(g) →C2H2 (g)

Find the heat of formation of Acetic acid at 25C                                            
2C + 2H2  +O2          CH3COOH                          
                                                         ∆H= ?
(a)    C  +  O2    →      CO2      ∆H=-394 kJl
(b)   H2  +  ½ O2  → H2O     ∆H=-286 kJ
(c)    CH3COOH  +  2O2   → 2CO2  +  2H2O                                             ∆H=-870.8 kJ



Explain Born –Haber cycle by using the enthalpy changes of the following reactions
Na(s) →Na(g)     ∆H=109 j/mol , Na(g) →Na+(g) ∆H= 494 j/mol
Cl2(g )→ 2Cl(g) ∆H=242 j/mol, Cl (g) →Cl-(g) ∆H= -363 j/mol and Na(s)+ ½ Cl2 (g)→ NaCl (s) ∆H= -471 j/mol

Calculate the enthalpy change  of                 Na+(g) +Cl-(g) →NaCl (s)

No comments:

Post a Comment