Question

Consider the reaction 4PH3(g) = P4(g) + 6H2(g)

Suppose that at a particular moment during the reaction, molecular hydrogen is being formed at the rate of 0.078 M/s.

(a) at what rate is P4 being formed?

(b) at what rate is PH3 reacting?

Answer 1

(a)

The rate of reaction $=\dfrac{d[P_4]}{dt}=\dfrac{d[H_2]}{6dt}$

Rate of P₄ changing $=\dfrac{d[P_4]}{dt}=\dfrac{d[H_2]}{6dt}=\dfrac{0.078}{6}=0.013M/s$

(b)The rate of reaction $= -\dfrac{d[PH_3]}{dt}=\dfrac{d[H_2]}{6dt}$

Rate of PH₃ changing $=-\dfrac{d[PH_3]}{dt}=4\dfrac{d[H_2]}{6dt}=\dfrac{4\times 0.078}{6}=0.052M/s$