Express the reverse equilibrium constant for the following reaction.

3 H2(g) + 3 Br2(g) 6 HBr(g)

Question

Express the reverse equilibrium constant for the following reaction.

3 H2(g) + 3 Br2(g)  6 HBr(g)

BlackTom AI · Accepted Answer

We are given the reaction: 3 H2(g) + 3 Br2(g) ⇌ 6 HBr(g) and asked for the reverse equilibrium constant.

First, recall how to write the equilibrium constant for a reaction in the forward direction. For aA + bB ⇌ cC, K_forward = [C]^c / ([A]^a [B]^b). Here, the forward reaction uses 3 H2 and 3 Br2 to form 6 HBr, so:
- A = H2 with a = 3
- B = Br2 with b = 3
- C = HBr with c = 6
Thus, K_forward = [HBr]^6 / ([H2]^3 [Br2]^3).

Now, the reverse reaction is the exact opposite: 6 HBr ⇌ 3 H2 + 3 Br2. The equilibrium constant for the reverse direction, K_reverse, is the reciprocal of the forward constant:
K_reverse = 1 / K_forward = ([H2]^3 [Br2]^3) / [HBr]^6.

To connect this to common answer formats, you can also think of K_reverse as:
K_reverse = ([H2]^3 [Br2]^3) / [HBr]^6 = ([H2]^3 [Br2]^3) × ([HBr]^{-6}).

Potential student pitfalls or misconceptions:
- Some students might mistakenly write the forward expression instead of the reciprocal for the reverse, which would give K_reverse = [HBr]^6 / ([H2]^3 [Br2]^3). Remember, the reverse constant is the reciprocal of the forward constant.
- Others might not apply the correct stoichiometric exponents. The exponents in the numerator correspond to the coefficients of the products in the reverse reaction (or the reactants in the forward reaction), and vice versa for the denominator.
- If options are written with simplified forms, make sure there’s no overlooked cancellation or factoring; the most compact and correct form here is the one with [H2]^3 [Br2]^3 in the numerator and [HBr]^6 in the denominator.

Because the provided answer options are not visible in the prompt, I cannot map these to specific letter choices. However, the correct conceptual form for the reverse equilibrium constant of the given reaction is K_reverse = ([H2]^3 [Br2]^3) / [HBr]^6, i.e., the reciprocal of the forward expression.

Express the reverse equilibrium constant for the following reaction. 3 H2(g) + 3 Br2(g) 6 HBr(g)

查看解析

登录即可查看完整答案

类似问题

Consider the reaction below: 3C(s) + 3H2(g) ⥫⥬ CH4(g) + C2H2(g) The equilibrium expression for the reaction above is:

Use the data given to calculate the value of Κ for the reaction at 5°C Ag+(aq) + Cl− (aq) AgCl(s) AgCl(s) Ag+(aq) Cl− (aq) S° (J K−1 mol−1) 96.2 72.68 56.4 ΔH°f (kJ/mol) −127.07 105.58 −167.2

Use the data given to calculate the value of Κ for the reaction at 25°C AgCl(s) Ag+(aq) + Cl− (aq) AgCl(s) Ag+(aq) Cl− (aq) S° (J K−1 mol−1) +96.2 +72.68 +56.4 ΔH°f (kJ/mol) −127.07 +105.58 −167.2

The equilibrium constant expression for the reaction shown below is Ag3PO4(s) 3 Ag+(aq) + PO43− (aq)

Express the equilibrium constant for the following reaction. 10 PCl5(g) 10 PCl3(g) + 10 Cl2(g)

Express the equilibrium constant for the following reaction. 7 N2(g) + 21 H2(g) ⇔ 14 NH3(g)

Express the equilibrium constant for the following reaction. CsIO3(s) CsIO(s) + O2(g)

Express the equilibrium constant for the following reaction. 3 P4(s) + 15 O2(g) 3 P4O10(s)

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单