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Difference between Internal Energy and Enthalpy

The enthalpy ...

Question

The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of $H_{2}$ at 1.5 atm pressure is $Δ H_{-}$ = -0.31 kJ. The value of $Δ E$ will be

-0.3024 kJ

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0.3024 kJ

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-2.567 kJ

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-0.0076 kJ

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Solution

The correct option is A
-0.3024 kJ

$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g) Δ n g = 1 - 2 = - 1; Δ H = - 0.31 K J m o l^{- 1} p = 1.5 a t m, Δ V = - 50 m L = - 0.050 L Δ H = Δ E + p Δ V - 0.31 = Δ E - 0.0076; Δ E = - 0.3024 K J$

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Similar questions

The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of $H_{2}$ at 1.5 atm pressure is $Δ H_{-}$ = -0.31 kJ. The value of $Δ E$ will be

Q. The enthalpy change for the reaction of

50

mL of ethylene with

50

mL of

H_{2}

1.5

atm pressure is

- 0.31 k J

. The value of

△ U

is:

Q. The enthalpy change for the reaction of

50 m l

of ethylene with

50.0 m l

H_{2}

1.5

atm pressure is

Δ H = - 0.31 K J

. What is the

Δ U

Q. The Enthalpy change for the reaction of

59 m l

of ethylene with

50.0 m l

H_{2}

1.5

bar pressure is $\Delta H=-0.31\ KJ$. What is the change in internal energy?

Q. The enthalpy change for the reaction

C_{2} H_{4} (g) + H_{2} (g) ⟶ C_{2} H_{6}

- 620 J

when 100ml of ethylene and 100 ml of

H_{2}

react. Calculate volume work and

Δ U .

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

The correct option is A -0.3024 kJ C2H4(g)+H2(g)→C2H6(g)Δng=1−2=−1;ΔH=−0.31KJmol−1p=1.5atm,ΔV=−50mL=−0.050LΔH=ΔE+pΔV−0.31=ΔE−0.0076;ΔE=−0.3024KJ

The enthalpy change for the reaction of 50.00 ml of ethylene with 50.00 ml of $H_{2}$ at 1.5 atm pressure is $Δ H_{-}$ = -0.31 kJ. The value of $Δ E$ will be

The correct option is A
-0.3024 kJ

$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g) Δ n g = 1 - 2 = - 1; Δ H = - 0.31 K J m o l^{- 1} p = 1.5 a t m, Δ V = - 50 m L = - 0.050 L Δ H = Δ E + p Δ V - 0.31 = Δ E - 0.0076; Δ E = - 0.3024 K J$