HOUSTON – Houston Texans QB C.J. Stroud has been named AFC Offensive Player of the Week, the NFL announced today.
This marks Stroud's third career Player of the Week honor following a standout performance in Houston's 44-10 win at Baltimore, where he completed 23-of-27 passes (85.2 percent) for 244 yards and four touchdowns, earning a 143.9 passer rating.
Stroud's four passing touchdowns tied for the most in the NFL in Week 5, while his passer rating ranked second across the league. His career-best 85.2 completion percentage also stood as the second highest in the NFL in Week 5 and the highest of his career (min. 20 attempts). Additionally, Stroud posted the longest rush of his career on a 30-yard run before capping that drive with a 10-yard touchdown pass to WR Nico Collins.
The third-year signal caller led a Texans offense that scored 44 points – tied for the third-most of any team in the NFL this season and the most points by any team in Week 5, the highest road point total in franchise history and helped deliver Houston's first-ever win at Baltimore.
Through five games this season, Stroud has completed 102-of-144 passes (70.8 percent) for 1,076 yards and eight touchdowns. Both his completion percentage and touchdown total rank fourth in the AFC. Over the last two weeks, Stroud has completed 45 of 55 passes (81.8 percent) for 477 yards, six touchdowns and a 139.2 passer rating — with both his completion percentage and passer rating leading the NFL during that span.
