Home
Class 12
CHEMISTRY
The enthalpy change for the reaction H(2...

The enthalpy change for the reaction `H_(2)(g)+C_(2)H_(4)(g)rarr C_(2)H_(6)(g)` is ……………… The bond energies are, `H - H=103, C-H=99, C-C=80` & `C=C=145 "K cal mol"^(-1)`

Promotional Banner

Similar Questions

Explore conceptually related problems

The enthalpy change for the reaction 2C("graphite")+3H_(2)(g)rarrC_(2)H_(6)(g) is called

The enthalpy change for the reaction 2C("graphite")+3H_(2)(g)rarrC_(2)H_(6)(g) is called

The enthalpy change for the reaction, C_(2)H_(6)(g)rarr2C(g)+6H(g) is X kJ. The bond energy of C-H bond is :

The enthalpy change for the reaction, C_(2)H_(6)(g)rarr2C(g)+6H(g) is X kJ. The bond energy of C-H bond is :

The enthalpy change for the reaction, C_(2)H_(6)(g)rarr2C(g)+6H(g) is X kJ. The bond energy of C-H bond is :

The enthalpy change for the reaction C_(2)H_(6(g)) rarr 2C_((g)) + 6H_((g)) is x kJ. The bond energy of C-H bond is:

Calculate enthalpy change of the following reaction : H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g)) The bond energy of C-H,C-C,C=C,H-H are 414,347,615 and 435k J mol^(-1) respectively.

Calculate enthalpy change of the following reaction : CH_(2) = CH_(2)(g) + H_(2)(g) rarr CH_(3) - CH_(3)(g) The bond energy of C - H, C - C, C = C, H - H are 414, 615 and 436 kJ mol^(-1) respectively.