At 448∘C, the equilibrium constant (K) for the reaction H2(g)+I2(g)<⇒2HI(g) is 50.5. If we start with 2.0×10−2 mol of HI,1.0×10−2mol of H2 and 3.0×10−2 mol of I2 in a 2 litre vessel, then at 448∘Ca. reaction is in equilibrium b. reaction will proceed in the forward direction c. reaction will proceed in the backward direction d. no reaction will take place.

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At {448}^{}{C}, the equilibrium constant (K) for the reaction

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At $448^{\circ} C$ , the equilibrium constant (K) for the reaction $H_{2 (g)} + I_{2 (g)} <\Rightarrow 2 H I_{(g)}$ is 50.5. If we start with $2.0 \times 10^{- 2}$ mol of $H I, 1.0 \times 10^{- 2}$ mol of $H_{2}$ and $3.0 \times 10^{- 2}$ mol of $I_{2}$ in a 2 litre vessel, then at $448^{\circ} C$ a. reaction is in equilibrium b. reaction will proceed in the forward direction c. reaction will proceed in the backward direction d. no reaction will take place.

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448^{\circ} C

, the equilibrium constant (K) for the reaction

H_{2 (g)} + I_{2 (g)} <\Rightarrow 2 H I_{(g)}

is 50.5. If we start with

2.0 \times 10^{- 2}

mol of

H I, 1.0 \times 10^{- 2}

mol of

H_{2}

and

3.0 \times 10^{- 2}

mol of

I_{2}

in a 2 litre vessel, then at

448^{\circ} C

a. reaction is in equilibrium b. reaction will proceed in the forward direction c. reaction will proceed in the backward direction d. no reaction will take place.

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Text solutionVerified

Sure, here are the step-by-step solutions:

Q = \frac{( 2.0 \times 10 ^{- 2} / 2 ) ^{2}}{( 1.0 \times 10 ^{- 3} / 2 ) ( 3.0 \times 10 ^{- 2} / 2 )}

Simplify the expression inside the numerator: {Q}=\frac{{{\left({1.0}\times{10}^{{-{2}}}}\right)}^{{2}}}{4{\left({1.0}\times{10}^{{-{3}}}\right)}{\left({3.0}\times{10}^{{-{2}}}\right)}} Simplify the expression inside the denominator: {Q}=\frac{{{\left({1.0}\times{10}^{{-{2}}}}\right)}^{{2}}}{6{\left({1.0}\times{10}^{{-{3}}}\right)}{\left({1.0}\times{10}^{{-{2}}}\right)}} Cancel out common factors: {Q}=\frac{{{\left({1.0}\times{10}^{{-{2}}}}\right)}}{6{\left({1.0}\times{10}^{{-{2}}}\right)}}=0.1667 Compare Q and K: Since

Q < K

, the reaction will proceed in the forward direction.

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Question Text	At $448^{\circ} C$ , the equilibrium constant (K) for the reaction $H_{2 (g)} + I_{2 (g)} <\Rightarrow 2 H I_{(g)}$ is 50.5. If we start with $2.0 \times 10^{- 2}$ mol of $H I, 1.0 \times 10^{- 2}$ mol of $H_{2}$ and $3.0 \times 10^{- 2}$ mol of $I_{2}$ in a 2 litre vessel, then at $448^{\circ} C$ a. reaction is in equilibrium b. reaction will proceed in the forward direction c. reaction will proceed in the backward direction d. no reaction will take place.
Topic	Chemical Kinetics
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1

Text solutionVerified

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