Answer
See the answer below.
Work Step by Step
a) Since the number of moles of reactants is equal to the number of moles of products, $K_p=K_c=56$
b) Initial concentrations: $0.045\ mol\div 10.0\ L=0.0045\ M$, reacted x
$K_c=[HI]^2/[H_2][I_2]$
$56=(2x)^2/(0.0045-x)^2$
$x=0.0036\ M$
Total pressure:
$P=(2\times0.0045)0.082\times698$
$P=0.515\ atm$
c) Partial pressures:
$p(HI)=0.515\ atm\times(2\times0.0036)/(2\times0.0045)=0.412\ atm$
$p(H_2)=p(I_2)=0.515\ atm\times(0.0009)/(2\times0.0045)=0.052\ atm$
