Question

Consider the following equilibrium.2 NOBr(g)<=> 2 NO(g) + Br2(g)If nitrosyl bromide, NOBr, is 34 percent dissociated at 25°C and the total pressure is 0.25 atm, calculate Kp and Kc for the dissociation at this temperature.

Answer 1

Answer:

$K_{p}$ of the reaction is  $9.521 \times 10^{-3}$.

$K_{c}$ of the reaction is  $3.89 \times 10^{-4}$.

Explanation:

Initial pressure of NOBr is "x" atm.

34% = 0.34

The equilibrium chemical reaction is a follows.

       $2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)$

Initial          x                                      0          0

Change    -0.34x                             +0.34x    +0.17x

Final        (x-0.34x)                          +0.34x     +0.17x

Total pressure = sum of the pressure at equilibrium.

$0.25atm=x-0.34x+0.34x+0.17x$

$0.25atm= 1.17x$

$x= \frac{0.25}{1.17}=0.213$

Partial pressure of NOBr=  x-0.34x =x (1-0.34)

= 0.213(0.66) = 0.14 atm.

Hence, partial pressure of $[]P_{NOBr}]$ is 0.14 atm.

$P_{NO}=0.34x = 0.34 \times 0.213 = 0.072$

$P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm$

From equation (1)

$K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}$

$= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}$

Therefore, $K_{p}$ of the reaction is  $9.521 \times 10^{-3}$.

$K_{p}= K_{c}(RT)^{\Delta n}$

Rearrange the equation is as follows.

$K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}$

$\Delta n$ = number of moles of reactants - Number of moles of products

= 3-2 = 1

$= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}$

Therefore,$K_{c}$ of the reaction is  $3.89 \times 10^{-4}$.

Answer 2

The $K_p$ and $K_c$ is " $0.009649615 \ atm$ and $0.00039\ mol\ L^{-1}$"for the dissociation at this temperature.

Equation:

$2NOBr\rightleftharpoons 2NO+Br_2$

​When the initial pressure of $NOBr$ is p, then at equilibrium, therefore the dissociation percentage of $NOBr \ \ is\ \ 34\%$

$\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p$

Calculating the total pressure:

$\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm$

$\to p=0.214 \ atm$

The equilibrium partial pressures:

$\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm$

Calculating the $K_p$:

$\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}$

          $=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm$

Calculating the $K_c$:

Using formula:

$\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}$

            $= (2+1) - 2\\\\= 3 - 2\\\\=1$

The relationship between $K_p$ and $K_c$ is:

$\to K_p = K_C(RT)^{\Delta ng}$

Hence, by putting the values of  $K_p$, gas constant $R, T,$ and $\Delta ng$, we can calculate the value of $K_c$.

$\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1$

$\to K_c = \frac{0.0096 }{0.0821\times 298}$

         $= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}$

Therefore, the $K_p$ and $K_c$ is " $0.009649615 \ atm$ and $0.00039\ mol\ L^{-1}$"for the dissociation at this temperature.

Find out more information about the Nitrosyl bromide here:

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