For the reaction SnO_{2(s)} + 2H_{2(g)} right | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$.Since $SnO_2$ and $Sn$ are in solid and liquid states respectiv

Vedclass · Accepted Answer

$1.49$

Vedclass · Answer

(A) The reaction is $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$.Since $SnO_2$ and $Sn$ are in solid and liquid states respectively,their activities are taken as $1$.Thus,the equilibrium constant $K_p$ is given by $K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2}$.Given that the mixture is $45\% \ H_2$ by volume,the mole fraction of $H_2$ is $0.45$ and the mole fraction of $H_2O$ is $1 - 0.45 = 0.55$.Since partial pressure is proportional to mole fraction,$K_p = \left( \frac{0.55}{0.45} \right)^2$.$K_p = (1.222)^2 \approx 1.49$.

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,calculate $K_p$ at $900 \ K$ where the equilibrium steam-hydrogen mixture was $45\% \ H_2$ by volume.

Similar Questions

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents

Which of the following is the correct order of increasing solubility for the given compounds?

$A$ liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference $P$. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ cut orthogonally,then the value of $k$ is

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module