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Question: The value of \({K_P}\) is \(1 \times {10^{ - 3}}{\text{ at}}{{\text{m}}^{ - 1}}\) at \({25^0}C\) for...

The value of KP{K_P} is 1×10−3 atm−11 \times {10^{ - 3}}{\text{ at}}{{\text{m}}^{ - 1}} at 250C{25^0}C for the reaction 2NO(g)+Cl2(g)⇆2NOCl2NO(g) + C{l_2}(g)\overset {} \leftrightarrows 2NOCl. A flask contains NO at 0.02 atm and at 250C{25^0}C. Calculate the mole of Cl2C{l_2} that must be added if 1% of the NO is to be converted to NOCl at equilibrium. The volume of the flask is such that 0.2 mole of gas produces 1 atm pressure at 250C{25^0}C. (Ignore probable association of NO in N2O2{N_2}{O_2})

Explanation

Solution

Finding the values of partial pressure from the formula of KP{K_P} as - KP=PproductsPreactants{K_P} = \dfrac{{{P_{products}}}}{{P_{reactants}^{}}}.
After we know the value in terms of partial pressure, we can convert in terms of moles using the ideal gas equation. The volume can be found from the volume of flask which is given.

Complete step by step answer:
For this, we will move step by step.
Let’s first write the reaction as –
2NO(g)+Cl2(g)⇆2NOCl2NO(g) + C{l_2}(g)\overset {} \leftrightarrows 2NOCl

From question, we have at time t =0 ; partial pressure of NO = 0.02atm
Further, we have to find moles of Cl2C{l_2} when 1% of NO is converted.
So, at equilibrium i.e., at t=tequ{t_{equ}} we are left with NO = 90% of 0.02
= 0.02×99100\dfrac{{0.02 \times 99}}{{100}} . Let P be the pressure of Cl2C{l_2} at equilibrium.
We know that 1% of the NO has been converted to NOCl at equilibrium. So, the pressure of NOCl is 0.02×1100\dfrac{{0.02 \times 1}}{{100}} .

Further, the formula for KP{K_P} is as - KP=PNOCLPNO2×PCl2{K_P} = \dfrac{{{P_{NOCL}}}}{{P_{NO}^2 \times {P_{C{l_2}}}}}
First, let us find the values of PNOCl{P_{NOCl}} and PNO2P_{NO}^2 as-

PNOCl{P_{NOCl}} = (0.02100)2{\left( {\dfrac{{0.02}}{{100}}} \right)^2} = 4×10−84 \times {10^{ - 8}}
PNO2P_{NO}^2 = (0.02×99100)2{\left( {\dfrac{{0.02 \times 99}}{{100}}} \right)^2} = 3.9×10−43.9 \times {10^{ - 4}}
Now, putting the values, we get - KP=4×10−83.9×10−4×P=1×10−3atm−1{K_P} = \dfrac{{4 \times {{10}^{ - 8}}}}{{3.9 \times {{10}^{ - 4}} \times P}} = 1 \times {10^{ - 3}}at{m^{ - 1}}
After this we can calculate the value of P = 0.102 atm

The value we got is in terms of partial pressure. But in the question we have been told to find moles.
For finding moles, we have ideal gas equation which is –
PV = nRT
We have P = 0.102 atm
R = gas constant
T = 250C{25^0}C = 298 K
V = Volume that we need to find from the values given in question.

First we will find Volume as –
For NO: we have been given that 0.2 mole of gas produces 1 atm pressure at 250C{25^0}C.
So, V=0.2×R×298V = 0.2 \times R \times 298.
Putting this value We have –

PV=nCl2RTPV = {n_{C{l_2}}}RT
0.102×0.2×R×298=nCl2×R×2980.102 \times 0.2 \times R \times 298 = {n_{C{l_2}}} \times R \times 298
nCl2=0.0204 mol{n_{C{l_2}}} = 0.0204{\text{ mol}}
Thus, we have 0.0204 moles and this is our answer.

Note: Here, we have used the KP{K_P} which is used when we write the concentrations at equilibrium in terms of partial pressure. It may be defined as the ratio of concentration of product and concentration of reactant in terms of partial pressure.
In case, the concentration is in terms of molar concentration, then we use KC{K_C} which is defined as the ratio of concentration of product and reactant in terms of molar concentration.
Do the numerical type questions step by step. Always mention the units to avoid confusion. Here, first we got the value of chlorine in terms of partial pressure which we have converted later on into moles.