The enthalpy change for the reaction of 50.00 mL of ethylene with 50.00 mL of H2 at 1.5 atm pressure is Δ H=-0.31 kJ. The value of Δ E will be:

Question

The enthalpy change for the reaction of 50.00 mL of ethylene with 50.00 mL of H2 at 1.5 atm pressure is Δ H=-0.31 kJ. The value of Δ E will be: (A) 0.235 kJ (B) 0.3024 kJ (C) 2.567 kJ (D) 0.0076 kJ

Tardigrade Admin · Accepted Answer

Use the following formula to find the value of

Δ E

.

Δ H - Δ E + P Δ V

where

Δ H = - 0.31 k J m o l^{- 1}

P = 1.5 a t m

Δ V = - 50 m L = - 0.050 L

Δ E = ?

Δ H = Δ E + P Δ V

or

- 0.31 = Δ E + (1.5 \times - 0.05)

or

- 0.31 = Δ E - 0.075

or

- 0.31 + 0.075 = Δ E

Δ E = 0.235

Q. The enthalpy change for the reaction of 50.00mL of ethylene with 50.00mL of H2​ at 1.5atm pressure is ΔH=−0.31kJ. The value of ΔE will be:

0.235 kJ

0.3024 kJ

2.567 kJ

0.0076 kJ

Use the following formula to find the value of ΔE. ΔH−ΔE+PΔV where ΔH=−0.31kJmol−1 P=1.5atm ΔV=−50mL=−0.050L ΔE=? ΔH=ΔE+PΔV or −0.31=ΔE+(1.5×−0.05) or −0.31=ΔE−0.075 or −0.31+0.075=ΔE ΔE=0.235

Q. The enthalpy change for the reaction of $50.00 m L$ of ethylene with $50.00 m L$ of $H_{2}$ at $1.5 a t m$ pressure is $Δ H = - 0.31 k J$ . The value of $Δ E$ will be: