Asked by Kaylyn
Calculate (delta) H rxn for
2NOCl(g)---- N2(g) + O2(g) + Cl(g)
given the following set of reactions:
1/2N2(g) + 1/2O2(g)-----NO(g) delta H =90.3kJ
NO(g) + 1/2Cl2(g)----NOCl(g) delta H = -38.6 kJ
2NOCl(g)---- N2(g) + O2(g) + Cl(g)
given the following set of reactions:
1/2N2(g) + 1/2O2(g)-----NO(g) delta H =90.3kJ
NO(g) + 1/2Cl2(g)----NOCl(g) delta H = -38.6 kJ
Answers
Answered by
DrBob222
Are you sure that equation 1 is correct? It isn't balanced. I think it should be
2NOCl ==> N2 + O2 + Cl2
To get that equation, multiply equation 2 by 2 and reverse it (multiply delta H by 2 and change the sign). Multiply equation 3 by 2 and reverse it (multiply delta H by 2 and change the sign). Then add the two equations to obtain the final equation. Add the multiplied delta Hs with the changed signs.
2NOCl ==> N2 + O2 + Cl2
To get that equation, multiply equation 2 by 2 and reverse it (multiply delta H by 2 and change the sign). Multiply equation 3 by 2 and reverse it (multiply delta H by 2 and change the sign). Then add the two equations to obtain the final equation. Add the multiplied delta Hs with the changed signs.
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