The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :

The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g)rarr N_(2)(g)+3H_(2)(g)` is :
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The enthalpy of formation of ammonia is `-46.0 KJ mol^(-1)` . The enthalpy change for the reaction `2NH_(3)(g) ... mol^(-1)` D. `-92.0 KJ mol^(-1)`
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The enthalpy change `(Delta H)` for the reaction `N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Description : The enthalpy change `(Delta H)` for the reaction `N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)` is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

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The value of enthalpy change `(DeltaH)` for the reaction `C_(2)H_(5)OH (l)+3O_(2) (g) rarr 2CO_(2) (g) +3H_(2)O (l)` at `27^(@)C` is `-1366.5 kJ mol^(

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Consider the reaction: `N_(2) + 3H_(2) hArr 2NH_(3)` carried out at constant pressure and temperature. If `DeltaH` and `DeltaU` are change in enthalpy

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The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

Description : The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)

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What is the effect of temperature and pressure on the yields of products? a. `N_(2)(s)+3H_(2)(g) hArr 2NH_(3)+x cal` b. `N_(2)(g)+O_(2)(g) hArr 2NO(g)

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For the given reaction , `A(g) rarr 2B(g) , Delta_(r)H= 30 kJ//"mole" Delta_(r)S = 150 J//mol` at `300 K` If `C_(P,A) = 20 J//K mol` and `C_(P,B) = 20

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For the reaction `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` the value of `Delta H=30.56 KJ mol^(_1)` and `Delta S = 66 JK^(-1)mol^(-1)`. The temperature at

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The temperature at which the reaction : `Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g)` is at equilibrium is …….., Given `Delta H=30.5 KJ mol^(-1)` and `Delta S

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Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite bu

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`P_(4)(s) + 3OH^(-) (aq) + 3H_(2)O (I) rarr PH_(3) (g) + 3H_(2)PO_(2)^(-) (aq)` In the above equation, the species getting oxidized and reduced respec

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Standard enthalpy of formation of ethane is -84.7kjper mol calculate the enthalpy change for the formation of 0.1kg ethane

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`Delta_(f)^(@)` for `CO_(2(g)), CO_((g))` and `H_(2)O_((g))` are `-393.5,-110.5` and `-241.8kJ mol^(-1)` respectively. The standard enthalpy change `(

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An element has successive ionization enthalpies as 940 (first),2080,3090,4140,7030,7870,16000 and 19500 kJ `mol^(-1)`. To which group of the periodic

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Standard free energy (∆G°) of hydrolysis of ATP to ADP + Pi is (A) –49.3 KJ/mol (B) –4.93 KJ/mol (C) –30.5 KJ/mol (D) –20.9 KJ/mol

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Steam is throttled to 0.1 MPa with 20 degrees of superheat. (a) What is the quality of throttled steam if its pressure is 0.75 MPa (b) What is the enthalpy of the process?  a) 97.6%,2713 kJ/kg  b) -97.6%, 2713 kJ/kg  c) 87.6%,3713 kJ/kg  d) -87.6%, 3713 kJ/kg

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In how many of the following reactions `N_(2)` gas may be released `{:((a) NH_(4)NO_(2) overset(Delta)rarr,(b) (NH_(4))_(2) Cr_(2)O_(7) overset(Delta)

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In how many of the following reactions `N_(2)` gas may be released `{:((a) NH_(4)NO_(2) overset(Delta)rarr,(b) (NH_(4))_(2) Cr_(2)O_(7) overset(Delta)

Description : In how many of the following reactions `N_(2)` gas may be released `{:((a) NH_(4)NO_(2) overset(Delta)rarr,(b) (NH_(4))_(2) Cr_(2)O_(7) overset(Delta)

Last Answer : In how many of the following reactions `N_(2)` gas may be released `{:((a) NH_(4)NO_(2) overset(Delta) ... i) Ba(N_(3))_(2) overset(Delta) rarr):}`

In the conversion `NH_(2)OH rarr N_(2)O`, the equivalent weight of `NH_(2)OH` will be:

Description : In the conversion `NH_(2)OH rarr N_(2)O`, the equivalent weight of `NH_(2)OH` will be:

Last Answer : In the conversion `NH_(2)OH rarr N_(2)O`, the equivalent weight of `NH_(2)OH` will be: A. M/4 B. M/2 C. M/5 D. M/1

In a boiler enthalpy of water supplied was 2000 kJ/kg. Enthalpy being added by fuel combustion is 3200 kJ/kg. Using first law of thermodynamics, find amount of heat supplied if steam generation rate is 10 tons per hour.

Description : In a boiler enthalpy of water supplied was 2000 kJ/kg. Enthalpy being added by fuel combustion is 3200 kJ/kg. Using first law of thermodynamics, find amount of heat supplied if steam generation rate is 10 tons per hour.

Last Answer : Heat supplied Q = m (h2-h1) …….(Steady flow energy equation for boiler) ; h2=3200 KJ/kg h1=2000 KJ/kg m = 10 Tons /hour = (10 x 1000)/3600 kg/sec = 2.77 Kg/sec Q = 2.77(3200-2000) = 3324 KJ/sec heat supplied for 10 Tons/hr steam generation

From the thermochemical reactions, C(graphite)`+1//2O_(2)rarr CO , Delta H = -110.5` KJ `CO+1//2O_(2)rarr CO_(2),Delta H=-283.2` KJ the heat of reacti

Description : From the thermochemical reactions, C(graphite)`+1//2O_(2)rarr CO , Delta H = -110.5` KJ `CO+1//2O_(2)rarr CO_(2),Delta H=-283.2` KJ the heat of reacti

Last Answer : From the thermochemical reactions, C(graphite)`+1//2O_(2)rarr CO , Delta H = -110.5` KJ `CO+1//2O_(2)rarr CO_( ... 7` KJ C. `-172.7` KJ D. `+172.7` KJ

The change in the enthalpy of `NaOH+HCl rarr NaCl+H_(2)O` is called :

Description : The change in the enthalpy of `NaOH+HCl rarr NaCl+H_(2)O` is called :

Last Answer : The change in the enthalpy of `NaOH+HCl rarr NaCl+H_(2)O` is called : A. Heat of ... of reaction C. Heat of hydration D. Heat of solution

Can anyone help me with this? Chlorine and carbon monoxide gases are mixed in a 1.00L container and the following equilibrium is established: CO(g) + Cl2(g) <--> COCl2(g).. Initially 1.5 mol of chlorine was present with excess carbon monoxide. At equilibrium, 0.8 mol of COCl2(g) was found. How do you calculate the percent reaction?

Description : Can anyone help me with this? Chlorine and carbon monoxide gases are mixed in a 1.00L container and the following equilibrium is established: CO(g) + Cl2(g) COCl2(g).. Initially 1.5 mol of ... monoxide. At equilibrium, 0.8 mol of COCl2(g) was found. How do you calculate the percent reaction?

Last Answer : For a start, your equation isn't balanced; it should be Cl2 (Cl2, not 2Cl, because of HONClBrIF). Given this, if we assume the reaction goes to completion, 1.5 mol Cl2 will produce 1.5 mol ... .3% Incidentally, is that supposed to be CO or Co? It has no effect on the calculation, just curious.