At 500 K, the equilibrium constant for the reaction H2(g) +I2(g)leftharpoons 2HI(g) is 24.8. If (1/2) textmol/L of HI is present at equilibrium, what are the concentrations of H2 and I2, assuming that we started by taking HI and reached the equilibrium at 500 K?

Question

At 500 K, the equilibrium constant for the reaction H2(g) +I2(g)leftharpoons 2HI(g) is 24.8. If (1/2) textmol/L of HI is present at equilibrium, what are the concentrations of H2 and I2, assuming that we started by taking HI and reached the equilibrium at 500 K? (A) 0.068 textmol L-1 (B) 1.020 textmol L-1 (C) 0.10 textmol L-1 (D) 1.20 textmol L-1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/cba13f75473241091c025a2b2d15528a-.png" /> Kc=([H2][I2]/[HI]2) or (1/24.8)=(xx/0.5×0.5) or x2=(0.25/24.8)=0.010 x=0.10 textmol L-1

$0.068 mol L^{- 1}$

$1.020 mol L^{- 1}$

$0.10 mol L^{- 1}$

$1.20 mol L^{- 1}$

$K_{c} = \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}}$ or $\frac{1}{24.8} = \frac{xx}{0.5 \times 0.5}$
or $x^{2} = \frac{0.25}{24.8} = 0.010$
$x = 0.10 mol L^{- 1}$

Q. At 500K, the equilibrium constant for the reaction H2(g)​+I2(g)​⇌2HI(g)​ is 24.8. If 21​mol/L of HI is present at equilibrium, what are the concentrations of H2​ and I2​, assuming that we started by taking HI and reached the equilibrium at 500K?

0.068mol L−1

1.020mol L−1

0.10mol L−1

1.20mol L−1

Kc​=[HI]2[H2​][I2​]​ or 24.81​=0.5×0.5xx​ or x2=24.80.25​=0.010 x=0.10mol L−1

$0.068 mol L^{- 1}$

$1.020 mol L^{- 1}$

$0.10 mol L^{- 1}$

$1.20 mol L^{- 1}$

$K_{c} = \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}}$ or $\frac{1}{24.8} = \frac{xx}{0.5 \times 0.5}$
or $x^{2} = \frac{0.25}{24.8} = 0.010$
$x = 0.10 mol L^{- 1}$