The equilibrium constant for the reaction H2(g) + I2(g) <=> 2HI(g) at a certain temperature is 6.0. What would be the equilibrium concentration of HI at that temperature, when the equilibrium concentration of H2 and I2 are 0.2 mol dm-3 and 0.3 mol dm-3 respectively?

Step 1: Write the equilibrium constant expression for the given reaction. The reaction is H_2(g) + I_2(g) 2HI(g). The equilibrium constant expression, K_c, is: K_c = [HI]^2[H_2][I_2] Step 2: Substitute the given values into the expression. We are given: K_c = 6.0 [H_2] = 0.2 mol dm^-3 [I_2] = 0.3 mol dm^-3 Substituting these values: 6.0 = [HI]^2(0.2 mol dm^-3)(0.3 mol dm^-3) Step 3: Calculate the denominator. (0.2)(0.3) = 0.06 (mol dm^-3)^2 So the equation becomes: 6.0 = [HI]^20.06 Step 4: Solve for [HI]^2. [HI]^2 = 6.0 × 0.06 [HI]^2 = 0.36 (mol dm^-3)^2 Step 5: Solve for [HI]. [HI] = sqrt(0.36 (mol dm)^-3)^2 [HI] = 0.6 mol dm^-3 Comparing this result with the given options, option D is 0.60 mol dm^-3. The final answer is D. 0.60 mol dm^-3. Send me the next one 📸