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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 $o4$.
$ OH = 30 kJ mol^{–1} , OS = 75 J / k / mol $ . Find boiling temperature at 1 atm.
To find the boiling temperature at 1 atm, use the Gibbs free energy equation at equilibrium: $\Delta G = \Delta H - T\Delta S = 0$. \n\nGiven: $\Delta H = 30\,kJ\,mol^{-1}$ and $\Delta S = 75\,J\,mol^{-1}\,K^{-1} = 0.075\,kJ\,mol^{-1}\,K^{-1}$.\n\nSet $\Delta G$ to zero and solve for $T$: \n$0 = 30\,kJ\,mol^{-1} - T(0.075\,kJ\,mol^{-1}\,K^{-1})$.\n\n$T = \frac{30\,kJ\,mol^{-1}}{0.075\,kJ\,mol^{-1}\,K^{-1}} = 400\,K$.\n\nThus, the boiling temperature at 1 atm is 400 K. The correct option is $o1$.
Spontaneous adsorption of a gas on a solid surface is exothermic process because
Spontaneous adsorption of a gas on a solid surface is generally an exothermic process. During adsorption, gas molecules adhere to the surface, resulting in a decrease in disorder (entropy) of the gas molecules because they are more confined. This leads to a decrease in the entropy of the system.\n\nTherefore, the correct option is $o3$, stating that entropy decreases during spontaneous adsorption.
The ratio of P to V at any instant is constant and is equal to 1, for a monoatomic ideal gas under going a process. What is the molar heat capacity of the gas
From first law of Thermodynamics, $ \triangle E = q + w \Rightarrow nC_vdT = nCdT – PdV...... (1) $
Now according to process, P = V and according to ideal gas equation, PV = nRT We have, V2 = nRT $ On differentiating, 2VdV = nRdT and PdV = VdV = { nRdT \over 2 } $
So from first equation we have, $ nC_vdT = nCdT – { nRdT \over 2 } $ So, $ C_v = C – { R \over 2 } $ $ Hence C = { 4R \over 2 } $
The entropy values (in J K–1 mol–1) of H2 (g) = 130.6 Cl2(g) = 223 and HCl(g) = 186.7 at 298 K and 1 atmpressure are given. Then entropy change for the reaction.
A mixture of 2 mole of CO(g) and one mole of O2 in a closed vessel, is ignited to convert the carbon monoxide to carbon dioxide. If $ \triangle H and \triangle U$ are enthalpy and internal energy change. Then
For the reaction of one mole zinc dust with one sulphuric acid in a bomb calorimeter, $ \triangle U $ and w correspond to :
In a bomb calorimeter, the volume is constant, so no work is done (w = 0). For the reaction of zinc with sulfuric acid: $$ ext{Zn(s) + H}_2 ext{SO}_4(aq) ightarrow ext{ZnSO}_4(aq) + ext{H}_2(g) $$, the internal energy change $$ riangle U $$ is negative because the reaction releases energy (exothermic reaction). Hence, the correct option is: $$ riangle U < 0, w = 0 $$.
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