Question 17.76: a) Write a reaction table given that PCH4 (init) = PCO2 (ini......
a) Write a reaction table given that \text{P}_{\text{CH}_4}\text{ (init) = P}_{\text{CO}_2}\text{ (init) = 10.0 atm}, substitute equilibrium values into the equilibrium expression, and solve for \text{P}_{\text{H}_2}.
b) Repeat the calculations for part (a) with the new K_\text{p} value. The reaction table is the same.
a)
A quadratic is necessary:
4\text{ x}^2 + (1.8836135 \times 10^3 \text{ x}) – 1.8836135 \times 10^4 = 0\text{ (unrounded)} \\ \text{a} = 4\quad\quad \text{b} = 1.8836135 \times 10^3\text{ c} = -1.8836135 \times 10^4 \\ \text{x}=\frac{-\text{b}\pm \sqrt{\text{b}^2-4\text{ac}} }{2\text{a}} \\ \text{x}=\frac{-1.8836135\times 10^3\pm \sqrt{(1.8836135\times 10^3)^2-4(4)(-1.8836135\times 10^4)} }{2(4)}
x = 9.796209 (unrounded)
P (hydrogen) = 2 x = 2 (9.796209) = 19.592418 atm (unrounded)
If the reaction proceeded entirely to completion, the partial pressure of \text{H}_2 would be 20.0 atm (pressure is proportional to moles, and twice as many moles of \text{H}_2 form for each mole of \text{CH}_4\text{ or CO}_2 that reacts).
b)
K_\text{p}=\frac{\text{P}_\text{CO}^2\text{ P}_{\text{H}_2}^2}{\text{P}_{\text{CH}_4}\text{ P}_{\text{CO}_2}} =\frac{(\text{2 x)}^2(\text{2 x)}^2}{(10.0-\text{x})(10.0-\text{x})} =\frac{(\text{2 x)}^4}{(10.0-\text{x})^2} =2.626\times 10^7 \\ \frac{(\text{2 x)}^2}{(10.0-\text{x})}=5.124451\times 10^3A quadratic is needed:
4\text{ x}^2+ (5.124451 \times 10^3 – 5.124451 \times 10^4=0 \text{ (unrounded)} \\ \text{a}=4\quad\quad \text{b} =5.124451 \times 10^3 \text{ c}=- 5.124451 \times 10^4 \\ \text{x}=\frac{-5.124451\times 10^3\pm \sqrt{(5.124451\times 10^3)^2-4(4)(-5.124451\times 10^4)} }{2(4)}
x = 9.923138 (unrounded)
P (hydrogen) = 2 x = 2 (9.923138) = 19.846276 atm (unrounded)
If the reaction proceeded entirely to completion, the partial pressure of \text{H}_2 would be 20.0 atm (pressure is proportional to moles, and twice as many moles of \text{H}_2 form for each mole of \text{CH}_4\text{ or CO}_2 that reacts).