For the reaction `N_(2)O_(5)(g)rarr 2NO_(2)(g)+(1)/(2)O_(2)(g)`, the value of rate of disappearance of `N_(2)O_(5)` is given as `6.5xx10^(-2)"mol L"^(-1)s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` is given respectively as

Explanation

Standard heat of formation of propane,
`3C_((s)) + 4H_(2(g)) to C_3H_(8(g)) " " DeltaH_f^0`=?
Data given :
`C_3H_(8(g)) + 5O_(2(g)) to 3CO_(2)+4H_2O_((l)) " " DeltaH="-2220.2 KJ mol"^(-1)`
`C_((s)) + O_(2(g)) to CO_(2(g)) " " DeltaH="-393.5 KJ mol"^(-1)`
`H_(2(g)) +1//2O_(2(g)) to H_2O_((l)) " " DeltaH=-"285.8 KJ mol"^(-1)`
According to Hess.s law , equation
Equation (1) is reversed
Equation (2) is x 3
Equation (3) is x 4
The add all the equations
`3CO_(2(g)) + 4H_2O_((l)) to C_3H_(8_(g)) +5O_(2(g)) " " DeltaH_1="+2220.2 KJ mol"^(-1)`
`3C_((s)) + 3O_(2(g)) to 3CO_(2(g)) " " DeltaH_2="-1180.5 KJ mol"^(-1)`
`4H_(2(g)) +2O_(2(g)) to 4H_2O_((l)) " " DeltaH_3="-1143.2 KJ mol"^(-1)`
`3C_((s)) + 4H_(2(g)) to C_3H_(8(g)) " " DeltaH_f^0="-103.5 KJ mol"^(-1)`
Standard enthalpy of formation of propane `DeltaH_f^0=-103.5 "KJ mol"^(-1)`