CH4(g) +Cl2(g) →CH3Cl (g) +HCl(g).has ∆H =-25 kCal
From the given data, what is the bond enthalpy of Cl-Cl bond
(A) 70 kCal
(B) 80 kCal
(C) 67.75 kCal
(D) 57.75 kCal
Correct option (D) 57.75 kCal
Explanation:
CH4 + Cl2 → CH3Cl + HCl
∆H = -25
–25 = 4 × C – H + 4Cl
– Cl – 3 × CH – 1C – Cl – 1 H – Cl
–25 = 4x + y – 3x – 84 – 103
y – x = 162