Question

For the reaction, $A\left(g\right)+2B\left(g\right)\rightleftharpoons 2C\left(g\right),K_C=1\times {10}^{-10}$ at ${25}^{\circ }C$. Which of the following statements is true?

• A. $△n=-1$
• B. The concentration of the products is greater than the concentration of the reactants
• C. The reaction is favoured in reverse direction
• D. The value of $K_p$ will be larger than the value of $K_C$

Answer 1

Answer: (A), (C)

The correct options are
A $△n=-1$
C The reaction is favoured in reverse direction

$A\left(g\right)+2B\left(g\right)\rightleftharpoons 2C\left(g\right),K_C=1\times {10}^{-10}$ at ${25}^{0}C$

$\Delta n=$ Difference in the no. of moles of gas between products and reactants

$=n_g^1-n_g^R$

(products) (reactants)

$=2-3$

$=-1$

Hence option $\left(A\right)$ is true

$A+2B\rightleftharpoons 2C$

$1$ $1$ $0$ (Initial concentration)

$1-x$ $1-2x$ $2x$ (At equilibrium)

$K_C=\frac{{\left(2x\right)}^2}{(1-x){(1-2x)}^2}=1\times {10}^{-10}$

$\approx 1$ $\approx 1$

Since $K_C<<1$ $x$ will be very small.

$\frac{{4x}^2}{\left(1\right)\left(1\right)}={10}^{-10}\Rightarrow x=1/2\times {10}^{-5}$

reactant concentration : $1-x+1-2x$

$=1-3x$ ( It can also be deduced logically

$=1-3/2\times {10}^{-5}$ Since $K_C<<1$ reaction hardly occurs).

Product concentration : $2x={10}^{-5}$

Clearly reactant concentration $>$ product concentration

Hence $\left(B\right)$ is false.

$K_P=K_C{\left(RT\right)}^{\Delta n}T={25}^{0}C=298F$ Here $K_C<1$

$K_P=(1\times {10}^{-10}){(0.0821\times 298)}^{-1}$ So reaction favoured reverse direction

$=\frac{{10}^{-10}}{0.0821\times 298}=\frac{{10}^{-10}}{24.46}$ Hence (C) also correct

Clearly $K_P<K_C$

Hence, $\left(D\right)$ is false.