0.5 mole of H 2( g ) and 1.0 mole of HI ( g ) (but no I 2 ) are added to a 1.0 L vessel and allowd to reach equilibrium according to the following reaction: H 2+ I 2 leftharpoons 2 HI ( g ) If x is the equilibrium concentration of I2, the correct equilibrium constant expression is

Question

0.5 mole of H 2( g ) and 1.0 mole of HI ( g ) (but no I 2 ) are added to a 1.0 L vessel and allowd to reach equilibrium according to the following reaction: H 2+ I 2 leftharpoons 2 HI ( g ) If x is the equilibrium concentration of I2, the correct equilibrium constant expression is (A) (x(0.5-x)/(1+2 x)2) (B) ((2-x)2/(0.5+x)2) (C) (x(2-x)2/(0.5-x)2) (D) ((1-2 x)2/(0.5+x) x)

Tardigrade Admin · Accepted Answer

underset0.5- x H 2+ underset1 I 2= underset1 - 2x2 HI ( g )

Q. $0.5$ mole of $H_{2} (g)$ and 1.0 mole of $H I (g)$ (but no $I_{2}$ ) are added to a $1.0 L$ vessel and allowd to reach equilibrium according to the following reaction: $H_{2} + I_{2} ⇌ 2 H I (g)$ If $x$ is the equilibrium concentration of $I_{2},$ the correct equilibrium constant expression is

$\frac{x ( 0.5 - x )}{( 1 + 2 x ) ^{2}}$

$\frac{( 2 - x ) ^{2}}{( 0.5 + x ) ^{2}}$

$\frac{x ( 2 - x ) ^{2}}{( 0.5 - x ) ^{2}}$

$\frac{( 1 - 2 x ) ^{2}}{( 0.5 + x ) x}$

$0.5 - x H_{2} + 1 I_{2} = 1 - 2 x 2 H I (g)$

Q. 0.5 mole of H2​(g) and 1.0 mole of HI(g) (but no I2​ ) are added to a 1.0L vessel and allowd to reach equilibrium according to the following reaction: H2​+I2​⇌2HI(g) If x is the equilibrium concentration of I2​, the correct equilibrium constant expression is

(1+2x)2x(0.5−x)​

(0.5+x)2(2−x)2​

(0.5−x)2x(2−x)2​

(0.5+x)x(1−2x)2​

0.5−xH2​​+1I2​​=1−2x2HI(g)​

$\frac{x ( 0.5 - x )}{( 1 + 2 x ) ^{2}}$

$\frac{( 2 - x ) ^{2}}{( 0.5 + x ) ^{2}}$

$\frac{x ( 2 - x ) ^{2}}{( 0.5 - x ) ^{2}}$

$\frac{( 1 - 2 x ) ^{2}}{( 0.5 + x ) x}$

$0.5 - x H_{2} + 1 I_{2} = 1 - 2 x 2 H I (g)$