Methane, ch4, reacts with i2 according to the reaction ch4(g)+i2(g)⇌ch3i(g)+hi(g)
at 630 k, kp for this reaction is 2.26×10−4. a reaction was set up at 630 k with initial partial pressures of methane of 105.1 torr and of 7.96 torr for i2.
calculate the pressure, in torr, of ch4.
express your answer to four significant figures and include the appropriate units.
calculate the pressure, in torr, of i2.
express your answer to three significant figures and include the appropriate units.
calculate the pressure, in torr, of ch3i.
express your answer to three significant figures and include the appropriate units.
calculate the pressure, in torr, of hi.
express your answer to three significant figures and include the appropriate units.

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

, where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr   7.96 torr  0       0             initial partial pressure

-x                  -x            +x     +x           react

105.1-x       7.96-x      x        x            equilibrium

Then:

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

the awnser is group 12