Asked by c chujor
write the equilibrium constant expression for each reaction a.) NH3(g)+ CIF3(g)-><3HF(g) +1/2N(g) +1/2Cl2(g)
Answers
Answered by
bobpursley
You need to use whole number coefficents to do this.
2NH3(g)+ 2CIF3(g)-><6HF(g) + N2(g) + Cl2(g)
Keq=[HF]<sup>6</sup>[N2][Cl2]/[NH3]<sup>2</sup>[ClF3]<sup>2</sup>
2NH3(g)+ 2CIF3(g)-><6HF(g) + N2(g) + Cl2(g)
Keq=[HF]<sup>6</sup>[N2][Cl2]/[NH3]<sup>2</sup>[ClF3]<sup>2</sup>
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