Δ H for the reaction, C (graphite) +2 H2(g) arrow C H4(g) at 298 K and 1 atm is -17900 cal. The Δ E for the above conversion would be

Question

Δ H for the reaction, C (graphite) +2 H2(g) arrow C H4(g) at 298 K and 1 atm is -17900 cal. The Δ E for the above conversion would be (A) − 17900 cal (B) 17900 cal (C) 17308 cal (D) − 17308 cal

Tardigrade Admin · Accepted Answer

C (graphite) +2 H 2(g) arrow CH 4(g) ; Δ H=-17900 cal Δ E=? Δ H=Δ E+Δ ng R T where, Δ ng= number of moles of gaseous products - number of moles of gaseous reactants =1-2=-1 ∴ Δ H=Δ E-1 × R × T -17900=Δ E-1 × 298 × 2 Δ E=-17900 +596 ∴ Δ E=-17304 cal

Q. ΔH for the reaction, C (graphite) +2H2(g)​→CH4(g)​ at 298K and 1atm is −17900cal. The ΔE for the above conversion would be

− 17900 cal

17900 cal

17308 cal

− 17308 cal

C (graphite) +2H2​(g)→CH4​(g);ΔH=−17900cal ΔE=?ΔH=ΔE+Δng​RT where, Δng​= number of moles of gaseous products - number of moles of gaseous reactants =1−2=−1 ∴ΔH=ΔE−1×R×T −17900=ΔE−1×298×2 ΔE=−17900+596 ∴ΔE=−17304cal

Q. $Δ H$ for the reaction, $C$ (graphite) $+ 2 H_{2 (g)} \to C H_{4 (g)}$ at $298 K$ and $1 a t m$ is $- 17900 c a l$ . The $Δ E$ for the above conversion would be