Question

for the reaction n2+3h2=2nh3 kp is 3.6x10¯² of 500k calculat kc valu

Answer 1

To find the relationship between \( K_p \) and \( K_c \), we can use the following formula:

\[
K_p = K_c (RT)^{\Delta n}
\]

where:
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in moles of gas.

For the reaction:

\[
N_2 + 3H_2 \rightleftharpoons 2NH_3
\]

We can find \( \Delta n \):
- \( n_{\text{products}} = 2 \) (from \( NH_3 \)),
- \( n_{\text{reactants}} = 1 + 3 = 4 \) (from \( N_2 + 3H_2 \)),
- Therefore, \( \Delta n = n_{\text{products}} - n_{\text{reactants}} = 2 - 4 = -2 \).

Given:
- \( K_p = 3.6 \times 10^{-2} \)
- \( T = 500 \, K \)

Now we can calculate \( K_c \):

1. Calculate \( RT \):
\[
RT = (0.0821 \, \text{L·atm/(K·mol)}) \times (500 \, K) = 41.05 \, \text{L·atm/(mol)}
\]

2. Now substitute these values into the equation for \( K_p \):
\[
K_p = K_c (RT)^{\Delta n}
\]
\[
3.6 \times 10^{-2} = K_c (41.05)^{-2}
\]

3. Calculate \( (41.05)^{-2} \):
\[
(41.05)^{-2} = \frac{1}{41.05^2} \approx \frac{1}{1686.6025} \approx 0.000592
\]

4. Substitute this back into the equation:
\[
3.6 \times 10^{-2} = K_c \times 0.000592
\]

5. Now, solve for \( K_c \):
\[
K_c = \frac{3.6 \times 10^{-2}}{0.000592} \approx 60.68
\]

Therefore, the value of \( K_c \) at 500 K is approximately \( 60.68 \).