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$ C (s) + O_2 (g) \rightarrow CO_2, (g)$ ; OH = –94.3 kcal/mol $ CO (g) + O_2(g) \rightarrow CO_2 (g)$ ; O H = – 67.4 kcal/mol $ O_2(g) \rightarrow 2O (g)$ ; OH = 117.4 kcal/mol $ CO (g) \rightarrow C (g) + O(g)$ ; OH = 230.6 kcal/mol Calculate OH for $ C (s) \rightarrow C (g) $ in kcal/mol.
The difference between OH and OE on a molar basis for the combustion of n–octane at $ 25 ^\circ C $ would be : $ 25 ^ \circ C $
$ \triangle H - \triangle E = -4.5 \times 8.315 \times 298 J = - 11.15 kJ $
What is the work done against the atmosphere when 25 grams of water vaporizes at 373 K against a constant external pressure of 1 atm ? Assume that steam obeys perfect gas laws. Given that the molar enthalpy of vaporization is 9.72 kcal/mole, what is the change of internal energy in the above process ?
In the reaction $ CS_2 (l) + 3O_2 (g) \rightarrow CO_2 (g) + 2SO_2 (g) OH = –265 kcal $ The enthalpies of formation of $ CO_2 and SO_2 $ are both negative and are in the ratio 4 : 3. The enthalpy of formation of $ CS_2 $ is + 26 kcal/mol. Calculate the enthalpy of formation of $ SO_2 $ .
$ Cs_2 (l) + 3 O_2 (g) \rightarrow Co_2 (g) + 2 SO_2 $ $ let \triangle H_f ( CO_2 , g ) = 4x and \triangle H_ f ( SO_2 , g ) = 3x $ $ \triangle H_{reaction} = \triangle H_f ( CO_2 , g ) = 2 \triangle H_f ( SO_2 .g ) - \triangle H_f ( CS_2 ) $ $ - 265 = 4x + 6x - 26 $ $ x = -23.9 $ $ \therefore \triangle H_f ( SO_2 , g ) = 3x = -71.7 Kcal / mol . $
The bond dissociation energy of gaseous $H_2, Cl_2 $ and HCl are 104, 58 and 103 kcal mol– 1 respectively. The enthalpy of formation for HCl gas will be
Given $ H_2 (g) \rightarrow HCl (g); \triangle H = 104 kcal ...(1) $ $ Cl_2 (g) \rightarrow 2Cl(g); \triangle H = 58 kcal ...(2) $ $ HCl (g) \rightarrow H(g) + Cl(g); \triangle H = 103 kcal ...(3) $ Heat of formation for HCl $ H_2 (g) + Cl_2 (g) \rightarrow HCl (g); \triangle H = ? $ Divide equation (1) and (2) by 2, and then add $ H_2 (g) + Cl_2 (g) \rightarrow H(g) + Cl(g); \triangle H = 81 kcal...(4) $ Subtracting equation (3) from equation (4) $ HCl (g) \rightarrow H(g) + Cl(g) ; \triangle H = 103 kcal ...(3) $ – – – –
$ H_2 (g) + Cl_2 (g) HCl(g); \triangle H = -22.0 kcal $ $ \therefore Enthalpy of formation of HCl gas = – 22.0 kcal $
$ AB, A_2 $ and $ B_2 $ are diatomic molecules. If the bond enthalpies of $A_2$, AB & $B_2 $ are in the ratio 1 : 1 : 0.5 and enthalpy of formation of AB from $A_2$ and $B_2$ is – 100 kJ/mol–1. What is the bond enthalpy of $A_2$ .
One mole of a gas occupying $3 dm^3 $ expands against constant external pressure of 1 atm to a volume of $ 13 dm^3 $ . The work done is –
The heat of formation of liquid methyl alcohol is kilojoule per mole using the following data will be [Heat of vaporisation of liquid methyl alcohol = 38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states : H, 218 kJ/mol; C, 715 kJ/mol; O, 249 kJ/mol. Average bond energies : C – H, 415 kJ/mol; C – O, 356 kJ/mol O – H, 463 kJ/mol.]
10 g of argon gas is compressed isothermally and reversibly at a temperature of $ 27 ^\circ C $ from 10 L to 5 L. q, W, OE and OH for this process are $ [R = 2.0 cal K^{–1} mol^{–1} , log_{10} 2 = 0.30] $ . [Atomic wt. of Ar = 40.]
Diborane is a potential rocket fuel which undergoes combustion according to the reaction, $ B_2H_6 (g) + 3O_2 (g) \rightarrow B_2O_3 (s) + 3H_2O(g) $ from the following data, the enthalpy change for the combustion of diborane will be $ 2B(s) + O_2 (g) \rightarrow B_2O_3(s) $ ; $ \triangle H $ = – 1273 kJ $ H_2(g) + O_2 (g) \rightarrow H_2O(l ) $ $ \triangle H $ = = – 286 kJ $ H_2O( l ) \rightarrow H_2O(g) $ ; $ \triangle H $ == 44 kJ $ 2B(s) + 2H_2 (g) \rightarrow B_2H_6 (g); $ $ \triangle H $ = = 46 kJ
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