For the reaction, H2F2(g) → H2(g) +F2(g) ; Δ E = - 14.2 K is cal/mole at 25°C . The change in enthalpy of the reaction is

Question

For the reaction, H2F2(g) → H2(g) +F2(g) ; Δ E = - 14.2 K is cal/mole at 25°C . The change in enthalpy of the reaction is (A)  13.6 kJ/mol  (B)  -13.6 kJ/mol  (C)  1.36 kJ/mol  (D)  -56.87 kJ/mol

Tardigrade Admin · Accepted Answer

Δ E = - 14.2 Kcal/mol = - 4.18 × 14.2 kJ/mol = - 59.356 kJ/mol ∵ Δ H = Δ E + p Δ V or Δ H = Δ E + Δ ng RT Δ H = - 59 .356 kJ /mol + 8.314 J/mol × 298 K [∴ Δng = 2 - 1 = 1 ] = - 59.356 kJ / mol + 2. 48 kJ /mol = - 56.87 kJ /mol

Q. For the reaction,
$H_{2} F_{2 (g)} \to H_{2 (g)} + F_{2 (g)}; Δ E = - 14.2 K$ is cal/mole at $2 5^{\circ} C$ . The change in enthalpy of the reaction is

$13.6 k J / m o l$

$- 13.6 k J / m o l$

$1.36 k J / m o l$

$- 56.87 k J / m o l$

Q. For the reaction, H2​F2(g)​→H2(g)​+F2(g)​;ΔE=−14.2K is cal/mole at 25∘C . The change in enthalpy of the reaction is

13.6kJ/mol

−13.6kJ/mol

1.36kJ/mol

−56.87kJ/mol

ΔE=−14.2Kcal/mol =−4.18×14.2kJ/mol =−59.356kJ/mol ∵ΔH=ΔE+pΔV or ΔH=ΔE+Δng​RT ΔH=−59.356kJ/mol+8.314J/mol×298K [∴Δng​=2−1=1] =−59.356kJ/mol+2.48kJ/mol =−56.87kJ/mol

Q. For the reaction,
$H_{2} F_{2 (g)} \to H_{2 (g)} + F_{2 (g)}; Δ E = - 14.2 K$ is cal/mole at $2 5^{\circ} C$ . The change in enthalpy of the reaction is

$13.6 k J / m o l$

$- 13.6 k J / m o l$

$1.36 k J / m o l$

$- 56.87 k J / m o l$