Solveeit Logo
Login

Question

Question: Iron fillings and water were placed in a 5 L vessel and sealed. The tank was heated to 1000°C. Upon ...

Iron fillings and water were placed in a 5 L vessel and sealed. The tank was heated to 1000°C. Upon analysis, the tank was found to contain 1.2 g of H₂(g) and 54.0 g of H₂O(g). If the reaction is represented as 3Fe(s)+4H2O(g)⇌Fe3O4(s)+4H2(g)3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g), then the value of equilibrium constant is

A

0.2

B

0.04

C

0.008

D

0.0016

Answer

0.0016

Explanation

Solution

The equilibrium constant expression is Kc=[H2]4[H2O]4K_c = \frac{[H_2]^4}{[H_2O]^4}. Moles of H2=1.2 g2 g/mol=0.6H_2 = \frac{1.2 \text{ g}}{2 \text{ g/mol}} = 0.6 mol. Moles of H2O=54.0 g18 g/mol=3.0H_2O = \frac{54.0 \text{ g}}{18 \text{ g/mol}} = 3.0 mol. Concentration of H2=0.6 mol5 L=0.12H_2 = \frac{0.6 \text{ mol}}{5 \text{ L}} = 0.12 M. Concentration of H2O=3.0 mol5 L=0.6H_2O = \frac{3.0 \text{ mol}}{5 \text{ L}} = 0.6 M. Kc=(0.120.6)4=(15)4=1625=0.0016K_c = \left(\frac{0.12}{0.6}\right)^4 = \left(\frac{1}{5}\right)^4 = \frac{1}{625} = 0.0016.