The concentration of OH ions in ammonia is 0.0090 M. Calculate the concentration

Question

The concentration of OH ions in ammonia is 0.0090 M. Calculate the concentration of H ions. Consider the following heterogenous equilibrium: CaCO_3(s) CaO_(s) + CO_2(g) At 800 degree C, the pressure of CO_2 is 0.783 Calculate K_p for the reaction at this temperature. Remember K_p = K_c(0.0821T)^Delta n For question number 2, calculate K_c. The initial concentrations of H_2, I_2, and HI are 0.00823 M, 0.00214 M, and 0.0624 M, respectively. The reaction is H_2(g) + I_2(g, 2HI_(g) K_c = 54.3 Calculate Q_c and determine the direction of the reaction

Explanation / Answer

1) the [OH-] in ammonia is 0.0090M

We kano in any solution [H+][OH-] = Kw = 1.0x10-14

Thus [H+] in the given solution =1.0x10-14 /0.009

= 1.11x10-12

2) CaCO3(s) <-----> CaO(s) + Co2(g)

Kp = PCO2 as [solids] = 1

thus Kp = PCo2 = 0.783

3) Kp = Kc {0.0821T]delta n

For the reaction in q2) delta n = 1-0 = 1

and Kp= 0.783

thus Kc = Kp/(0.0821x 1073)

= 0.783/(0.0821x 1073)

= 8.8x10-3

4) H2 + I2 <----> 2HI

Kc = [HI]2/[H2][I2]

Qc also will have the same form as Kc.

Thus Qc for the reaction at given concentrations is

Qc = (0.0624)2/(0.00823)(0.00214)

= 221.08

As Qc is much larger than Kc = 54.3, the direction of the reaction is toward left / backwards

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