Standard molar enthalpy of formation- Delta-f HV is just a special case of enthalpy of reaction- Delta-r HV - Is the Delta-r HV for the following reaction same as Delta-f HV - Give reason for your answer- CaO-s- - CO2-g- rarr- CaCO3-s- Delta-f HV - ndash-178-3 kJ mol ndash-1 23- The value of Delta-f HV for NH3 is ndash- 91-8 kJ mol ndash-1- Calculate enthalpy change for the following reaction - 2NH3-g- rarr- N2-g- - 3H2-g-

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Standard XII

Chemistry

Depression in Freezing Point

Standard mola...

Question

Standard molar enthalpy of formation, Δf HV is just a special case of enthalpy of reaction, Δr HV . Is the Δr HV for the following reaction same as Δf HV ? Give reason for your answer. CaO(s) + CO2(g) → CaCO3(s); Δf HV = –178.3 kJ mol–1 23. The value of Δf HV for NH3 is – 91.8 kJ mol–1. Calculate enthalpy change for the following reaction : 2NH3(g) → N2(g) + 3H2(g)

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Solution

$H e r e t h e m o l a r e n t h a l p y o f f o r m a t i o n o f C a C O_{3} i s g i v e n f o r i t s s \tan d a r d c o n d i t i o n a n d i s t h e r e a c t i o n w h i c h r e p r e s e n t s f o r 1 m o l e . C a O + C O_{2} \to C a C O_{3} S o ∆ H_{f}^{0} = - 178.3 k J / m o l e = ∆ H_{r} i i) 2 N H_{3} \to N_{2} + 3 H_{2} G i v e n d a t a i s f o r t h e f o r m a t i o n o f 1 m o l e o f a m m o n i a ∆ H_{f}^{0} = - 91.8 k J / m o l e H e r e t h e r e q u i r e d e q u a t i o n i s d e c o m p o s i t i o n o f a m m o n i a w h i c h i s t h e r e v e r s e . T h a t i s ∆ H_{r}^{0} = + 91.8 k J / m o l e S i n c e i t s f o r 2 m o l e a m m o n i a ∆ H_{r} = 2 \times 91.8 = 183.6 k J / m o l e$

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Standard molar enthalpy of formation, $Δ_{f} H^{⊖}$ is just a special case of enthalpy of reaction, $Δ_{r} H^{⊖}$ . Is the $Δ_{r} H^{⊖}$ for the following reaction same as $Δ_{f} H^{⊖}$ ? Give reason for your answer.
$C a 0 (s) + C O_{2} (g) \to C a C O_{3} (s); Δ_{f} H^{⊖} = - 178.3 k J m o l^{- 1}$