Question

Convert the values of Kc to values of Kp or the values of Kp to values of Kc.
A) N2(g)+3H2(g) <--> 2NH3(g); Kc=0.50 at 400 degrees Celsius.
B) H2+I2 <---> 2HI; Kc= 50.2 at 448 degrees Celsius.
C) Na2SO4*10H2O(s) <---> Na2SO4(s)+10H2O(g). Kp=4.08x10^-25 at 25 degrees Celsius.
D) H2O(l) <---> H2O (g); Kp= 0.122 at 50 degrees Celsius.

Answer 1

Answer:

For A: The value of $K_p$ for the given equation is $1.64\times 10^{-4}$

For B: The value of $K_p$ for the given equation is 50.2

For C: The value of $K_c$ for the given equation is $5.312\times 10^{-39}$

For D: The value of $K_c$ for the given equation is $4.60\times 10^{-3}$

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$         ..........(1)

where,

$K_p$ = equilibrium constant in terms of partial pressure

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}

$

T = temperature

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}$

• For A:

The given chemical equation follows:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

We are given:

$K_c=0.50\\T=400^oC=[400+273]K=673K\\\Delta n_g=2-4=-2$

Putting values in equation 1, we get:

$K_p=0.50\times (0.0821\times 673)^{-2}\\\\K_p=1.64\times 10^{-4}$

Hence, the value of $K_p$ for the given equation is $1.64\times 10^{-4}$

• For B:

The given chemical equation follows:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

We are given:

$K_c=50.2\\T=448^oC=[448+273]K=721K\\\Delta n_g=2-2=0$

Putting values in equation 1, we get:

$K_p=50.2\times (0.0821\times 721)^{0}\\\\K_p=50.2$

Hence, the value of $K_p$ for the given equation is 50.2

• For C:

The given chemical equation follows:

$Na_2SO_4.10H_2O(s)\rightleftharpoons Na_2SO_4(s)+10H_2O(g)$

We are given:

$K_p=4.08\times 10^{-25}\\T=25^oC=[25+273]K=298K\\\Delta n_g=10-0=10$

Putting values in equation 1, we get:

$4.08\times 10^{-25}=K_c\times (0.0821\times 298)^{10}\\\\K_c=5.312\times 10^{-39}$

Hence, the value of $K_c$ for the given equation is $5.312\times 10^{-39}$

• For D:

The given chemical equation follows:

$H_2O(l)\rightleftharpoons H_2O(g)$

We are given:

$K_p=0.122\\T=50^oC=[50+273]K=323K\\\Delta n_g=1-0=1$

Putting values in equation 1, we get:

$0.122=K_c\times (0.0821\times 323)^{1}\\\\K_c=4.60\times 10^{-3}$

Hence, the value of $K_c$ for the given equation is $4.60\times 10^{-3}$

Answer 2

Kp and Kc are the equilibrium constant. Kp for A. $1.64 \times 10^{-4}$, for B. 50.2 and Kc for C is $5.312 \times 10^{-39}$ and D. $4.60\times 10^{-3}.$

What are Kp and Kc?

Kp is the equilibrium constant given relative to the partial pressure whereas, Kc is given relative to the concentration. The relation between Kp and Kc can be shown as:

$\rm K_{p} = K_{c} (RT)^{\Delta\;ng}$

For reaction A the balanced reaction is shown as:

$\rm N_{2}(g)+3H_{2}(g) \leftrightharpoons 2NH_{3}(g)$

The value of Kp is:

$\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\&= 0.50\times (0.0821\times 673)^{-2}\\\\&= 1.64\times 10^{-4}\end{aligned}$

Thus, the Kp for A is $1.64\times 10^{-4}.$

For reaction B the balanced reaction is shown as:

$\rm H_{2} + I_{2} \rightleftharpoons 2HI$

The value of Kp is:

$\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\&= 50.2\times (0.0821\times 721)^{0}\\\\&= 50.2 \end{aligned}$

Thus, the Kp for B is 50.2.

For reaction C the balanced reaction is shown as:

$\rm Na_{2}SO_{4} .10H_{2}O(s) \rightleftharpoons Na_{2}SO_{4}(s)+10H_{2}O(g)$

The value of Kc is:

$\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\4.08\times 10^{-25} &= \rm K_{c} \times (0.0821\times 298)^{10}\\\\\rm K_{c} &= 5.312\times 10^{-39} \end{aligned}$

Thus, the Kc for C is $5.312\times 10^{-39}.$

For reaction D the balanced reaction is shown as:

$\rm H_{2}O(l) \rightleftharpoons H_{2}O$

The value of Kc is:

$\begin{aligned} \rm K_{p} &= \rm K_{c} (RT)^{\Delta ng}\\\\0.122 &= \rm K_{c} \times (0.0821\times 323)^{1}\\\\\rm K_{c} &= 4.60\times 10^{-3} \end{aligned}$

Thus, the Kc for D is $4.60\times 10^{-3}.$

Learn more about Kp and Kc here:

https://brainly.com/question/4615606