Arrange the following in order of increasing tendency of the forward reactions to proceed towards completion at `298 K` and one atmospheric pressure :
a. `H_(2)O(g)hArrH_(2)O(l), K_(c)=782`
b. `F_(2)(g)hArr2F(g), K_(c)=4.9xx10^(-21)`
c. `C_("graphite")+O_(2)(g)hArrCO_(2)(g), K_(c)=1.3xx10^(69)`
d. `N_(2)O_(4)(g)hArr2NO_(2)(g), K_(c)=4.6xx10^(-3)`
e. `H_(2)(g)+C_(2)H_(4)(g)hArrC_(2)H_(6)(g), K_(c)=9.8xx10^(18)`

Explanation

To solve the problem of arranging the reactions in order of increasing tendency of the forward reactions to proceed towards completion at 298 K and one atmospheric pressure, we will analyze the equilibrium constants (Kc) provided for each reaction. The larger the Kc value, the greater the tendency for the forward reaction to proceed towards completion.

### Step-by-Step Solution:

1. **Identify the Kc values** for each reaction:
- a. \( H_2O(g) \rightleftharpoons H_2O(l), K_c = 782 \)
- b. \( F_2(g) \rightleftharpoons 2F(g), K_c = 4.9 \times 10^{-21} \)
- c. \( C_{(graphite)} + O_2(g) \rightleftharpoons CO_2(g), K_c = 1.3 \times 10^{69} \)
...