H 2 ( g ) + B r 2 ( g ) → ← 2 H B r ( g ) k c = 2.18 × 1 0 6 T = 730 ° C H_{2(g)} + Br_{2(g)} \ ^\leftarrow_\rightarrow \ 2HBr_{(g)}\\
k_c = 2.18 × 10^6 \\
T = 730°C H 2 ( g ) + B r 2 ( g ) → ← 2 H B r ( g ) k c = 2.18 × 1 0 6 T = 730° C
number of moles of HBr = 3.20 mol
V = 12.0 L
molarity of HBr = 3.2/12 = 0.27M
Using Ice,
H + Br = 2HBr
I ---- 0M + 0M = 0.27M
C ----- +x +x = -2x
E ----- x + x = 0.27 -2x
k c = [ H B r ] 2 [ H 2 ] [ B r 2 ] 2.18 × 1 0 6 = [ 0.27 − x ] 2 [ x ] [ x ] 2.18 × 1 0 6 = ( 0.27 − x x ) 2 2.18 × 1 0 6 = 0.27 − x x 1476.5 = 0.27 − x x 1476.5 x = 0.27 − x 1477.5 x = 0.27 x = 0.000183 M \begin{aligned}
k_c &= \dfrac{[HBr]^2}{[H_2][Br_2]}\\
\\
2.18 × 10^6 &= \dfrac{[0.27-x]^2}{[x][x]}\\
\\
2.18 × 10^6 &= (\dfrac{0.27-x}{x})^2\\
\\
\sqrt{2.18 × 10^6} &= \dfrac{0.27-x}{x}\\
\\
1476.5 &= \dfrac{0.27-x}{x}\\
\\
1476.5x &= 0.27 - x\\
1477.5x &= 0.27\\
x &= 0.000183M
\end{aligned} k c 2.18 × 1 0 6 2.18 × 1 0 6 2.18 × 1 0 6 1476.5 1476.5 x 1477.5 x x = [ H 2 ] [ B r 2 ] [ H B r ] 2 = [ x ] [ x ] [ 0.27 − x ] 2 = ( x 0.27 − x ) 2 = x 0.27 − x = x 0.27 − x = 0.27 − x = 0.27 = 0.000183 M
∴ [ H 2 ] = x = 1.83 × 1 0 4 M [ B r 2 ] = x = 1.83 × 1 0 4 M [ H B r ] = 0.27 − 2 x = 0.27 − 2 ( 0.000183 ) = 0.27 M \begin{aligned}
\therefore [H_2] &= x = 1.83 × 10^4M \\
[Br_2] &= x = 1.83 × 10^4M\\
[HBr] &= 0.27 - 2x = 0.27 - 2(0.000183) = 0.27M
\end{aligned} ∴ [ H 2 ] [ B r 2 ] [ H B r ] = x = 1.83 × 1 0 4 M = x = 1.83 × 1 0 4 M = 0.27 − 2 x = 0.27 − 2 ( 0.000183 ) = 0.27 M
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