1. Tardigrade
  2. Question
  3. Chemistry
  4. The enthalpy and entropy change for the reaction textB ( textr)2 (. textl .) + textC ( textl)2 (. textg .) arrow 2 textB textr textC textl (. textg .) are 30 textk textJ text text textm texto textl - 1 texta textn textd 105 textJ textK - 1 text text textm texto textl - 1 respectively. The temperature at which the reaction will be in equilibrium is

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Q. The enthalpy and entropy change for the reaction
$\text{B} \left(\text{r}\right)_{2} \left(\right. \text{l} \left.\right) + \text{C} \left(\text{l}\right)_{2} \left(\right. \text{g} \left.\right) \rightarrow 2 \text{B} \text{r} \text{C} \text{l} \left(\right. \text{g} \left.\right)$
are $30 \, \text{k} \text{J} \, \text{} \text{} \text{m} \text{o} \text{l}^{ - 1} \, \, \text{a} \text{n} \text{d} \, \, 105 \, \text{J} \text{K}^{ - 1} \, \text{} \text{} \text{m} \text{o} \text{l}^{ - 1}$ respectively. The temperature at which the reaction will be in equilibrium is

NTA AbhyasNTA Abhyas 2022

Solution:

We know that
$\text{} \text{} \text{Δ} \text{G} = \text{} \text{} \text{Δ} \text{H} - \text{T} \text{Δ} \text{S}$
When the reaction is in equilibrium, $\text{} \text{} \text{Δ} \text{G} = 0$
$0=\text{}\text{}\text{Δ}\text{H}-\text{T}\text{Δ}\text{S}\Rightarrow \text{T}=\frac{\text{Δ} \text{H}}{\text{} \text{Δ} \text{S}} \\ \text{T}=\frac{30 \times 1000}{105}=285.7 \, \text{K}$