The reaction: H2(g) + I2(g) = 2HI(g) is a second order reaction ?

Question

(first order with respect to each of [H2(g)] and [I2(g)]). When [H2] = 0.200 mol/L and [I2] = 0.200 mol/L the rate is -d[H2]/dt = 0.00097 mol/L/s at 327oC. What will -d[H2]/dt be when [H2] = 0.400 mol/L and [I2] = 0.100 mol/L?

Answer 1

first order with respect to each of [H2(g)] and [I2(g]

if we double H2 ,...& cut I2 in half,...what happens to the rate?

your answer is : it stays the same @  -d[H2]/dt = 0.00097 mol/L/s