The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :

NO :

Cl2 :

![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)
Answer:
-2.86x10³ kJ
Explanation:
The enthalpy of a reaction (ΔH) is defined as the heat produced or consumed by a reaction. In the reaction:
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)
The ΔH is the heat envolved in the reaction per 2 moles of C₂H₆. 1.43x10³ kJ are involved when 1 mole reacts. Thus, when 2 moles react, involved heat is:
1.43x10³ kJ ₓ 2 = <em>2.86x10³ kJ</em>. As the reaction is a combustion reaction (Produce CO₂ and H₂O), the heat involved in the reaction is <em>PRODUCED, </em>that means ΔH is negative, <em>-2.86x10³ kJ</em>