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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
A sample of argon gas at 1 atm pressure and $ 27 ^\circ C $ expands reversibly and adiabatically from $ 1.25 dm^3 to 2.50 dm^ 3 $ . The enthalpy change in this process will be……….$ [Cv.m. for argon is 12.48 jK^{–1} mol^{–1} ]$.
Find $ OG ^\circ $ and $ OH ^\circ $ for that the reaction $ CO(g) + O_2 (g) \rightarrow CO_2 (g) $ at 300 K respectively are, when the standard entropy change is $ – 0.094 kJ mol^{–1} K^{–1} $ . The standard Gibbs free energies of formation for $ CO_2 and CO are – 394.4 and – 137.2 kJ mol^{–1} $ , respectively.
To solve for $\Delta G^\circ$ and $\Delta H^\circ$, we use the Gibbs free energy change formula: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $T = 300\,K$, $\Delta S^\circ = -0.094\,kJ\,mol^{-1}\,K^{-1}$, $ ext{standard Gibbs free energies of formation: } \Delta G^\circ_{CO_2} = -394.4\,kJ\,mol^{-1}$ and $\Delta G^\circ_{CO} = -137.2\,kJ\,mol^{-1}$.\n\nFirst, calculate $\Delta G^\circ$ for the reaction: $\Delta G^\circ = (-394.4\,kJ\,mol^{-1}) - (-137.2\,kJ\,mol^{-1}) = -257.2\,kJ\,mol^{-1}$.\n\nNow, use the Gibbs free energy formula to find $\Delta H^\circ$: \n$-257.2\,kJ\,mol^{-1} = \Delta H^\circ - (300\,K)(-0.094\,kJ\,mol^{-1}\,K^{-1})$.\n\nSolving for $\Delta H^\circ$: \n$\Delta H^\circ = -257.2\,kJ\,mol^{-1} + 28.2\,kJ\,mol^{-1} = -285.4\,kJ\,mol^{-1}$.\n\nThus, the correct values are $\Delta G^\circ = -257.2\,kJ\,mol^{-1}$ and $\Delta H^\circ = -285.4\,kJ\,mol^{-1}$. The correct option is $o4$.
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