C_ (dia) +O_2 CO_2 ; Δ H=-395.4 kJ / mole C_ (gr ) + O_2 CO_2 ; Δ H=-393.5 kJ / mole C_ (gr ) C_ (dia) ; Δ H= ?

Correct answer: D. +1.9 C( dia )+O_2 arrow CO_2(Δ H=-395.4 ~kJ / mole) ...(1) C(gr)+O_2 CO_2(Δ H=-393.5 ~kJ / mole) ...(2) subtracting eq. (1) from (2), we have aligned & C(gr) C_ (dia) ; & aligned (Δ H & =+1.9 ~kJ / mole) & [ Δ H=-393.5-(-395.4) & .=+1.9 ~kJ ~mol^ -1 ] aligned aligned

JIPMER 2005

Exam: JIPMER
Year: 2005
Subject: Chemistry
Chapter: Thermodynamics (C)

JIPMER 2005 · Chemistry · Thermodynamics (C)

C_ (dia) +O_2 CO_2 ; Δ H=-395.4 kJ / mole C_ (gr ) + O_2 CO_2 ; Δ...

\mathrm{C}_{\text{(dia) }}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H=-395.4 \mathrm{ kJ} / \mathrm{mole}

\mathrm{C}_{(\mathrm{gr}

)}+

\mathrm{O}_2

\longrightarrow

\mathrm{CO}_2

; \Delta H=-393.5

\mathrm{ kJ}

/

\mathrm{mole} \mathrm{C}_{(\mathrm{gr}

)} \longrightarrow

\mathrm{C}_{\text{(dia) }

} ; \Delta H=

?

Options

A-3.8
B-1.9
C+3.8
D+1.9

Correct answer

D. +1.9

Step-by-step solution

\mathrm{C}(

dia

)+\mathrm{O}_2 \rightarrow \mathrm{CO}_2(\Delta \mathrm{H}=-395.4 \mathrm{~kJ} / \mathrm{mole})

...(1)

\mathrm{C}(\mathrm{gr})+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2(\Delta \mathrm{H}=-393.5 \mathrm{~kJ} / \mathrm{mole})

...(2) subtracting eq. (1) from (2), we have

\begin{aligned} & \mathrm{C}(\mathrm{gr}) \longrightarrow \mathrm{C}_{\text{(dia) }} ; \\ & \begin{aligned}(\Delta \mathrm{H} & =+1.9 \mathrm{~kJ} / \mathrm{mole}) \\ & \quad[\because \Delta \mathrm{H}=-393.5-(-395.4) \\ & \left.=+1.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\right]\end{aligned}

\end{aligned}$

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