Question 6 of 69beginner🔧 ApplyNumerical2 marks

For the following equilibrium, Kc=6.3×1014K_c = 6.3 \times 10^{14} at 1000 K

NO(g)+O3(g)⇌NO2(g)+O2(g)NO (g) + O_3 (g) \rightleftharpoons NO_2 (g) + O_2 (g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is KcK_c, for the reverse reaction?

Correct Answer

KcK_c for the reverse reaction is 1.587×10−151.587 \times 10^{-15}

Exercise: EXERCISES | Q: 6.6 | (Page 41)
For More Understanding

Explanation

For any reversible reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Given the forward reaction NO(g)+O3(g)⇌NO2(g)+O2(g)NO (g) + O_3 (g) \rightleftharpoons NO_2 (g) + O_2 (g) has Kc=6.3×1014K_c = 6.3 \times 10^{14}, the reverse reaction NO2(g)+O2(g)⇌NO(g)+O3(g)NO_2 (g) + O_2 (g) \rightleftharpoons NO (g) + O_3 (g) will have an equilibrium constant equal to 1/Kc1 / K_c.

Solution Steps

  1. Step 1: Identify the relationship between the equilibrium constants of the forward and reverse reactions. The equilibrium constant for the reverse reaction (Kc(reverse)K_{c(reverse)}) is the reciprocal of the equilibrium constant for the forward reaction (Kc(forward)K_{c(forward)}).

    Kc(reverse)=1Kc(forward)K_{c(reverse)} = \frac{1}{K_{c(forward)}}

  2. Step 2: Substitute the given value of the forward equilibrium constant into the equation.

    Kc(forward)=6.3×1014K_{c(forward)} = 6.3 \times 10^{14} Kc(reverse)=16.3×1014K_{c(reverse)} = \frac{1}{6.3 \times 10^{14}}

  3. Step 3: Calculate the final value.

    Kc(reverse)=1.587×10−15K_{c(reverse)} = 1.587 \times 10^{-15}
PreviousNext