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Question: The equilibrium constant for the following reaction is \( 1.6 \times {10^5} \) at 1024K. \( {H_{...

The equilibrium constant for the following reaction is 1.6×1051.6 \times {10^5} at 1024K.
H2(g)+Br2(g)⇌2HBr(g){H_{2(g)}} + B{r_{2(g)}} \rightleftharpoons 2HB{r_{(g)}}
Find the equilibrium pressure of all gases if 10.0bar10.0bar of HBr is introduced into a sealed container at 1024K.

Explanation

Solution

Hint : The equilibrium constant Kp{K_p} gives us a relationship between the partial pressures of reactants and products and is calculated using the partial pressures of the reaction equation. It is Unit less quantity.

Complete Step By Step Answer:
We are given the Kp{K_p} value for the reaction H2(g)+Br2(g)⇌2HBr(g){H_{2(g)}} + B{r_{2(g)}} \rightleftharpoons 2HB{r_{(g)}} which is 1.6×1051.6 \times {10^5} . The Kp{K_p} values for the inverse reaction 2HBr⇌H2(g)+Br2(g)2HBr \rightleftharpoons {H_{2(g)}} + B{r_{2(g)}} Kp′K{'_p} will be the inverse of Kp{K_p}
Therefore we can say that Kp′=1Kp=11.6×105=6.25×10−6K{'_p} = \dfrac{1}{{{K_p}}} = \dfrac{1}{{1.6 \times {{10}^5}}} = 6.25 \times {10^{ - 6}}
We are given that the initial pressure is 10.0bar10.0bar . Let ‘p’ be the change in pressure at final/equilibrium change. The reaction can be give as:
2HBr⇌H2(g)+Br2(g)2HBr \rightleftharpoons {H_{2(g)}} + B{r_{2(g)}}
Initial 10 - -
Final 10−2p10 - 2p pp pp
The value of Kp′K{'_p} can be given as: Kp′=pH2×pBr2pHBr2K{'_p} = \dfrac{{{p_{{H_2}}} \times {p_{B{r_2}}}}}{{p_{HBr}^2}}
Kp′=p×p(10−2p)2=6.25×10−6K{'_p} = \dfrac{{p \times p}}{{{{(10 - 2p)}^2}}} = 6.25 \times {10^{ - 6}}
p10−p=2.5×10−3\dfrac{p}{{10 - p}} = 2.5 \times {10^{ - 3}}
p=2.5×10−2−(5.0×10−3)pp = 2.5 \times {10^{ - 2}} - (5.0 \times {10^{ - 3}})p
p+(5.0×10−3)p=2.5×10−2p + (5.0 \times {10^{ - 3}})p = 2.5 \times {10^{ - 2}}
p=2.49×10−2bar≈2.5×10−2barp = 2.49 \times {10^{ - 2}}bar \approx 2.5 \times {10^{ - 2}}bar
The partial pressures of Hydrogen and Bromine gas is equal to p, hence the partial pressures of
pH2=pBr2=p=2.5×10−2bar{p_{{H_2}}} = {p_{B{r_2}}} = p = 2.5 \times {10^{ - 2}}bar
pHBr=10−2p=10−2(2.5×10−2)≈10bar{p_{HBr}} = 10 - 2p = 10 - 2(2.5 \times {10^{ - 2}}) \approx 10bar (Since the value of 2p is very small we can neglect it)
Therefore the partial pressures of all the gases are found.
Remember that in an arithmetic equation if the value (comparatively larger) is added or subtracted from a value ⩽0.05\leqslant 0.05 that value can be neglected, since it will cause negligible change in the overall answer.

Note :
Always remember that the equilibrium constant of an inverse reaction (inverse to the one given in the question) will be the reciprocal of the original equilibrium constant. This is applicable for both Kp&Kc{K_p}\& {K_c} . For example consider the following reaction:
A+B⇌ABA + B \rightleftharpoons AB Kp{K_p} Kc{K_c}
The equilibrium constant for the reverse reaction, AB⇌A+BAB \rightleftharpoons A + B denoted by Kc′K{'_c} can be given as:
Kc′=1KcK{'_c} = \dfrac{1}{{{K_c}}}
The terms Kp&Kc{K_p}\& {K_c} are related to each other by a simple equation, Kp=Kc(RT)Δn{K_p} = {K_c}{(RT)^{\Delta n}}
Where, Δn\Delta n is the difference in the moles of product and reactant. T is the temperature and R is the gas constant.