For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ \(K_{\mathrm{p}}=3.5 \times 10^{4}\) at \(1495 \mathrm{~K}\). What is the value of \(K_{\mathrm{p}}\) for the following reactions at \(1495 \mathrm{~K}\) ? a. \(\mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\) b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\)

Short Answer

Expert verified
The values of \(K_p\) for the following reactions at \(1495\) K are: a. \(K_{p(a)} = \left(\frac{1}{3.5 \times 10^4}\right)^{\frac{1}{2}} \approx 5.67 \times 10^{-3}\) b. \(K_{p(b)} = \frac{1}{3.5 \times 10^4} \approx 2.86 \times 10^{-5}\) c. \(K_{p(c)} = \left(3.5 \times 10^4\right)^{\frac{1}{2}} \approx 187\)

Step by step solution

01

Recall Equilibrium Constant Manipulation Rules

When manipulating a balanced chemical equation, we need to remember two main rules concerning the equilibrium constant: - If the reaction is reversed, the new equilibrium constant is the reciprocal of the original one. - If the balanced equation is multiplied or divided by a scalar (n), then the new equilibrium constant is raised to the power of (n). 2. Find the value of \(K_p\) for Reaction a
02

Calculate K_p for Reaction a

In reaction a, the given reaction is reversed and divided by 2. This means we need to take the reciprocal of the given \(K_p\) and raise it to the power of (1/2). \[ K_{p(a)} = \left(\frac{1}{K_p}\right)^{\frac{1}{2}} \] Substitute the given \(K_{p}\) value: \[ K_{p(a)} = \left(\frac{1}{3.5 \times 10^4}\right)^{\frac{1}{2}} \] Now calculate the value for \(K_{p(a)}\). 3. Find the value of \(K_p\) for Reaction b
03

Calculate K_p for Reaction b

Reaction b is simply the reverse of the given reaction. This means we need to take the reciprocal of the given \(K_p\). \[ K_{p(b)} = \frac{1}{K_p} \] Substitute the given \(K_{p}\) value: \[ K_{p(b)} = \frac{1}{3.5 \times 10^4} \] Now calculate the value for \(K_{p(b)}\). 4. Find the value of \(K_p\) for Reaction c
04

Calculate K_p for Reaction c

In reaction c, the given reaction is divided by 2. This means we need to raise the given \(K_p\) to the power of (1/2). \[ K_{p(c)} = \left(K_p\right)^{\frac{1}{2}} \] Substitute the given \(K_{p}\) value: \[ K_{p(c)} = \left(3.5 \times 10^4\right)^{\frac{1}{2}} \] Now calculate the value for \(K_{p(c)}\).

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Most popular questions from this chapter

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Explain the difference between \(K, K_{\mathrm{p}}\), and \(Q\).

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