For a reaction A2+B2 leftharpoons 2AB the figure shows the path of the reaction in absence and presence of a catalyst. What will be the energy of activation for forward (Ef) and backward (Eb) reaction in presence of a catalyst and Δ H for the reaction? The dotted curve is the path of reaction in presence of a catalyst.

Question

For a reaction A2+B2 leftharpoons 2AB the figure shows the path of the reaction in absence and presence of a catalyst. What will be the energy of activation for forward (Ef) and backward (Eb) reaction in presence of a catalyst and Δ H for the reaction? The dotted curve is the path of reaction in presence of a catalyst. (A) Ef =60 KJ /mol, Eb=70KJ /mol, Δ H=20 KJ /mol (B) Ef =20 KJ /mol, Eb=20KJ /mol, Δ H=50 KJ /mol (C) Ef =70 KJ /mol, Eb=20KJ /mol, Δ H=10 KJ /mol (D) Ef =10 KJ /mol, Eb=20KJ /mol, Δ H=10 KJ /mol

Tardigrade Admin · Accepted Answer

Ef =70-60=10 kJ/ mol Eb =70-50=20 kJ/ mol Δ H =50-60 =-10 kJ /mol

$E_{f} = 60 K J / m o l$ , $E_{b} = 70 K J / m o l$ , $Δ H = 20 K J / m o l$

$E_{f} = 20 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 50 K J / m o l$

$E_{f} = 70 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 10 K J / m o l$

$E_{f} = 10 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 10 K J / m o l$

$E_{f}$ $= 70 - 60 = 10 k J / m o l$
$E_{b}$ $= 70 - 50 = 20 k J / m o l$
$Δ H$ $= 50 - 60$ $= - 10 k J / m o l$

Ef​=60KJ/mol, Eb​=70KJ/mol, ΔH=20KJ/mol

Ef​=20KJ/mol, Eb​=20KJ/mol, ΔH=50KJ/mol

Ef​=70KJ/mol, Eb​=20KJ/mol, ΔH=10KJ/mol

Ef​=10KJ/mol, Eb​=20KJ/mol, ΔH=10KJ/mol

Ef​ =70−60=10kJ/mol Eb​ =70−50=20kJ/mol ΔH =50−60 =−10kJ/mol

$E_{f} = 60 K J / m o l$ , $E_{b} = 70 K J / m o l$ , $Δ H = 20 K J / m o l$

$E_{f} = 20 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 50 K J / m o l$

$E_{f} = 70 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 10 K J / m o l$

$E_{f} = 10 K J / m o l$ , $E_{b} = 20 K J / m o l$ , $Δ H = 10 K J / m o l$

$E_{f}$ $= 70 - 60 = 10 k J / m o l$
$E_{b}$ $= 70 - 50 = 20 k J / m o l$
$Δ H$ $= 50 - 60$ $= - 10 k J / m o l$