Login
Get Advice on Live Video Call
Earn $ Cash $ with
consultations on Bissoy App

Equimolar mixture of two gases \( A_{2} \) and \( B_{2} \) is taken in a rigid vessel at temperature \( 300 K \). The gases reacts according to given equations : \[ \begin{array}{ll} A_{2}(g) \rightleftharpoons 2 A(g) & K_{P_{1}}=? \\ B_{2}(g) \rightleftharpoons & K_{P_{2}}=? \\ A_{2}(g)+B_{2}(g) & \rightleftharpoons 2 A B(g) & K_{P_{3}}=2 \end{array} \] If the initial pressure in the container was 2 atm and final pressure developed at equilibrium is \( 2.75 atm \). in which equilibrium partial pressure of gas \( A B \) was \( 0.5 atm \), calculate the ratio of \( \frac{ K _{ p _{2}}}{ K _{ p _{1}}} \). [Given : Degree of dissociation of \( B_{2} \) is greateı than \( A _{2} \) ].

Monjur Alahi

Equimolar mixture of two gases \( A_{2} \) and \( B_{2} \) is taken in a rigid vessel at temperature \( 300 K \). The gases reacts according to given equations : \[ \begin{array}{ll} A_{2}(g) \rightleftharpoons 2 A(g) & K_{P_{1}}=? \\ B_{2}(g) \rightleftharpoons & K_{P_{2}}=? \\ A_{2}(g)+B_{2}(g) & \rightleftharpoons 2 A B(g) & K_{P_{3}}=2 \end{array} \] If the initial pressure in the container was 2 atm and final pressure developed at equilibrium is \( 2.75 atm \). in which equilibrium partial pressure of gas \( A B \) was \( 0.5 atm \), calculate the ratio of \( \frac{ K _{ p _{2}}}{ K _{ p _{1}}} \). [Given : Degree of dissociation of \( B_{2} \) is greateı than \( A _{2} \) ].


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

this is the solution!!!! 

0 like 0 dislike
1 views
4. The value of \( \left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} \) is (a) \( \frac{61}{144} \) (b) \( \frac{144}{61} \) (c) 29 (d) none of these

Correct answer is:- (c) 29 Explanation:-  1/2-2 + 1/3-2 + 1/4-2 = 22 + 32 + 42 (x-1 = 1/x) = 4 + 9 + 16  = 29 

2 Answers 1 views
12 In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by \( \hat{ k } \) and \( 2 \hat{ i }-2 \hat{ j } \), respectively. What is the unit vector along direction of propagation of the wave? (a) \( \frac{1}{\sqrt{2}}(\hat{ i }+\hat{ j }) \) (b) \( \frac{1}{\sqrt{2}}(\hat{ j }+\hat{ k }) \) (c) \( \frac{1}{\sqrt{5}}(\hat{ i }+2 \hat{ j }) \) (d) \( \frac{1}{\sqrt{5}}(2 \hat{ i }+\hat{ j }) \)

Correct answer is (a) We know that propagation wave vector ∵ \(\vec{E} = \hat{k}\) \(\vec{B} = 2\hat{i} - 2\hat{j}\) \(\vec{C} = \vec{E} \times \vec{B}\) \(\vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1...

2 Answers 1 views
equilibrium constant is The value of \( K_{ p } \) for the equilibrium \( NH _{4} H S_{(s)} \rightleftharpoons NH _{3(g)}+ H _{2} S _{(g)} \) at a certain temperature and total 2 atm pressure at equilibrium will be \( atm ^{2} \)

We have given, Total pressure at equilibrium = 2 atm. Constant temperature = NH4HS(s) \(\rightleftharpoons\)NH3(g) + H2S(g) Kp = \(\frac{P_{NH_3}\times P_{H_2S}}{P_{NH_4HS}}\) ∵ Partial pressure of solid substance is considered to be 1. ∴ Kp = \(P_{NH_3}\times P_{H_2S}\)  Kp = \((P_{NH_3})^2\) (From the...

2 Answers 1 views
Consider a particle of mass \( m \) having linear momentum \( P \) at position \( r \) relative to the origin \( O \). Which of the following equations correctly relates \( r , P , L \) ? a) \( \left[\left(\frac{d L}{d t}\right)+r x\left(\frac{d P}{d t}\right)\right]=0 \) b) \( \left[\left(\frac{d L}{d t}\right)-r x\left(\frac{d P}{d t}\right)\right]=0 \) c) \( \left[\left(\frac{d L}{d t}\right) \times\left(\frac{d r}{d t}\right) \times P\right]=0 \) d) \( \left[\left(\frac{d L}{d t}\right)-\left(\frac{d r}{d t}\right) \times P\right]=0 \)

Correct answer is (b) We know that \(\tau\) = r × F ∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\) \(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\) \(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)

2 Answers 1 views
If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \)

b^2 -4ac = 0 (as roots are equal ) b^2 =4ac b^2/4a = c

2 Answers 1 views
If the roots of the equation \( a x^{2}+b x+c=0 \) are equal then \( c=? \) \( \quad \) 1) \( -\frac{b}{2 a} \) 2) \( \frac{b}{2 a} \) 3) \( -\frac{b^{2}}{4 a} \) 4) \( \frac{b^{2}}{4 a} \) 2 Answers 1 views
\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0

1/1+n^3 +4/n^3+1 put n equal to 1/0  then 1/1+{1/0}^3 0/1 then answer is D

2 Answers 1 views
\( \lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}+1}+\frac{4}{n^{3}+1}+\frac{9}{n^{3}+1}+\ldots .+\frac{n^{2}}{n^{3}+1}\right) \) is equal to (A) 1 (B) \( 2 / 3 \) (C) \( 1 / 3 \) (D) 0 2 Answers 1 views
(1.) If \( T={ }_{B}^{A}\left(\begin{array}{cc}0.25 & 0.75 \\ 0.3 & \alpha\end{array}\right) \) is a transition probability matrix, then the value of \( \alpha \) is a) \( 0.5 \) b) \( 0.7 \) c) 0 /d) \( 0.6 \) 2. Rank of matrix \( \left(\begin{array}{cccc}1 & -1 & 2 & 0 \\ 0 & 2 & 0 & 3\end{array}\right) \) is a) 1 b) 0 c) 2 d) \( -1 \)

Is there any answer

2 Answers 2 views
(1.) If \( T={ }_{B}^{A}\left(\begin{array}{cc}0.25 & 0.75 \\ 0.3 & \alpha\end{array}\right) \) is a transition probability matrix, then the value of \( \alpha \) is a) \( 0.5 \) b) \( 0.7 \) c) 0 /d) \( 0.6 \) 2. Rank of matrix \( \left(\begin{array}{cccc}1 & -1 & 2 & 0 \\ 0 & 2 & 0 & 3\end{array}\right) \) is a) 1 b) 0 c) 2 d) \( -1 \) 2 Answers 1 views

Log in to ask questions, provide answers, or leave comments.


Login

Install the Bissoy app to consult with a doctor.


Download