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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 |
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Answer» Solution :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)` |
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