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Given reaction: H2(g) + Br2(g) 2HBr(g) with K_p = 7.1 x 104 at 700 K.

Chemistry

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Answers

2 parts
a

1.408×10−5\mathbf{1.408 \times 10^{-5}}

b

266.458\mathbf{266.458}

Step-by-step solution

4 steps

Given reaction: H2(g)+Br2(g)⇌2HBr(g)H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g) with Kp=7.1×104K_p = 7.1 \times 10^4 at 700 K.

Part a· 2 steps

For the reaction 2HBr(g)⇌H2(g)+Br2(g)2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)

  1. Identify the relationship between the given reaction and the target reaction.

    The target reaction is the reverse of the original reaction.

  2. Calculate the new KpK_p.

    When a reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant.

    Kp′=1KpK_p' = \frac{1}{K_p} Kp′=17.1×104K_p' = \frac{1}{7.1 \times 10^4} Kp′=171000K_p' = \frac{1}{71000} Kp′≈1.408×10−5K_p' \approx 1.408 \times 10^{-5}

    The value of KpK_p is 1.408×10−5\boxed{\mathbf{1.408 \times 10^{-5}}}.

Part b

266.458\mathbf{266.458}

For the reaction 12H2(g)+12Br2(g)⇌HBr(g)\frac{1}{2}H_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons HBr(g)

  1. Identify the relationship between the given reaction and the target reaction.

    The target reaction is the original reaction multiplied by a factor of 12\frac{1}{2}.

  2. Calculate the new KpK_p.

    When a reaction is multiplied by a factor 'n', the new equilibrium constant is the original constant raised to the power 'n'. Here, n=12n = \frac{1}{2}.

    Kp′′=(Kp)1/2=KpK_p'' = (K_p)^{1/2} = \sqrt{K_p} Kp′′=7.1×104K_p'' = \sqrt{7.1 \times 10^4} Kp′′=71000K_p'' = \sqrt{71000} Kp′′≈266.458K_p'' \approx 266.458

    The value of KpK_p is 266.458\boxed{\mathbf{266.458}}.

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