2 NF 3 (g) + Cu(s) --> N 2 F 4 (g) +CuF 2 (s) H = ? using the following thermoch

Question

2 NF3(g) + Cu(s) --> N2F4(g) +CuF2(s)     H = ? using the following thermochemical equations     NF3(g) + NO(g) --> 1/2N2F4(g) + ONF(g)    H =-41.4 kJ     2 NO(g) + F2(g) --> 2ONF(g)                       H = -313.8 kJ     Cu(s) + F2(g) -->CuF2(s)                              H = -513.0 kJ 2 NF3(g) + Cu(s) --> N2F4(g) +CuF2(s)     H = ? using the following thermochemical equations     NF3(g) + NO(g) --> 1/2N2F4(g) + ONF(g)    H =-41.4 kJ     2 NO(g) + F2(g) --> 2ONF(g)                       H = -313.8 kJ     Cu(s) + F2(g) -->CuF2(s)                              H = -513.0 kJ

Explanation / Answer

This question is easily answered using Hess' law which statedthat the total enthalpy change can be aquired byadding intermediate reactions together of known change ofEnthaply.
2*(NF3(g) + NO(g) --> 1/2N2F4(g) + ONF(g)    H =-41.4 kJ) -1*(2 NO(g) + F2(g) --> 2ONF(g)                       H = -313.8 kJ) 1*( Cu(s) + F2(g) -->CuF2(s)                              H = -513.0 kJ) +________________________________________________________ = 2 NF3(g) + Cu(s) --> N2F4(g) +CuF2(s)     H = ? ( aNegative equation means products are no reactants andreactants are now products) H = 2*-41KJ -1*-313.3KJ *1-513.0KJ = -282KJ

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